OP Malhotra Class 12 Maths Solutions Chapter 19 Bayes Theorem Exercise 19

 

Question 1. Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B.
Answer: Let's define the following events:
\( E_1 \): Bag A is chosen.
\( E_2 \): Bag B is chosen.
\( A \): A red ball is drawn.

Since there are two bags and one is chosen at random, the probabilities of choosing each bag are equal:
\( P(E_1) = \frac{1}{2} \)
\( P(E_2) = \frac{1}{2} \)

Now, we find the probability of drawing a red ball from each bag:
\( P(A | E_1) = \text{Probability of drawing a red ball from bag A} \)
Bag A has 2 white and 3 red balls, so a total of \( 2+3=5 \) balls.
\( P(A | E_1) = \frac{3}{5} \)

\( P(A | E_2) = \text{Probability of drawing a red ball from bag B} \)
Bag B has 4 white and 5 red balls, so a total of \( 4+5=9 \) balls.
\( P(A | E_2) = \frac{5}{9} \)

We want to find the probability that the red ball was drawn from bag B, which is \( P(E_2 | A) \). We use Bayes' Theorem:
\( P(E_2 | A) = \frac{P(E_2) P(A | E_2)}{P(E_1) P(A | E_1) + P(E_2) P(A | E_2)} \)

\( \implies P(E_2 | A) = \frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}} \)

\( \implies P(E_2 | A) = \frac{\frac{5}{18}}{\frac{3}{10} + \frac{5}{18}} \)

\( \implies P(E_2 | A) = \frac{\frac{5}{18}}{\frac{3 \times 9 + 5 \times 5}{90}} \)

\( \implies P(E_2 | A) = \frac{\frac{5}{18}}{\frac{27+25}{90}} \)

\( \implies P(E_2 | A) = \frac{\frac{5}{18}}{\frac{52}{90}} \)

\( \implies P(E_2 | A) = \frac{5}{18} \times \frac{90}{52} \)

\( \implies P(E_2 | A) = \frac{5 \times 5}{52} \)

\( \implies P(E_2 | A) = \frac{25}{52} \)
In simple words: First, we listed the different choices for bags and drawing a red ball. Then, we used a special rule called Bayes' Theorem. This rule helps us find out the chance that a red ball came from Bag B, knowing that a red ball was picked. The final probability is 25 out of 52.

🎯 Exam Tip: Clearly define your events \( E_1, E_2, A \) and their probabilities at the start. Mistakes often happen in the common denominator step of Bayes' Theorem, so be careful with calculations.

 

Question 2. There are two bags I and II, containing 3 red and 4 white balls, and 2 red and 3 white balls respectively. A bag is selected at random and a ball is drawn from it. It is found to be a red ball, find the probability that it is drawn from the first bag.
Answer: Let's define the events as follows:
\( E_1 \): Bag I is selected.
\( E_2 \): Bag II is selected.
\( E \): A red ball is drawn.

Since a bag is selected at random from two bags:
\( P(E_1) = \frac{1}{2} \)
\( P(E_2) = \frac{1}{2} \)

Now, we find the probability of drawing a red ball from each bag:
\( P(E | E_1) = \text{Probability of drawing a red ball from bag I} \)
Bag I contains 3 red and 4 white balls, so a total of \( 3+4=7 \) balls.
\( P(E | E_1) = \frac{3}{7} \)

\( P(E | E_2) = \text{Probability of drawing a red ball from bag II} \)
Bag II contains 2 red and 3 white balls, so a total of \( 2+3=5 \) balls.
\( P(E | E_2) = \frac{2}{5} \)

We want to find the probability that the red ball was drawn from Bag I, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{\frac{3}{7} \times \frac{1}{2}}{\frac{3}{7} \times \frac{1}{2} + \frac{2}{5} \times \frac{1}{2}} \)

\( \implies P(E_1 | E) = \frac{\frac{3}{14}}{\frac{3}{14} + \frac{2}{10}} \)

\( \implies P(E_1 | E) = \frac{\frac{3}{14}}{\frac{3}{14} + \frac{1}{5}} \)

\( \implies P(E_1 | E) = \frac{\frac{3}{14}}{\frac{15+14}{70}} \)

\( \implies P(E_1 | E) = \frac{\frac{3}{14}}{\frac{29}{70}} \)

\( \implies P(E_1 | E) = \frac{3}{14} \times \frac{70}{29} \)

\( \implies P(E_1 | E) = \frac{3 \times 5}{29} \)

\( \implies P(E_1 | E) = \frac{15}{29} \)
In simple words: We used the given numbers for red and white balls in two bags. We also know that picking either bag is equally likely. Since we picked a red ball, we wanted to find the chance that it came from the first bag. Bayes' Theorem helped us combine these facts to get the answer.

🎯 Exam Tip: Always make sure to calculate the total number of balls in each bag correctly before finding the probability of drawing a specific color ball. Simplify fractions early to make calculations easier.

 

Question 3. Suppose that 5 men out of 100 and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is chosen. Assume that there are equal number of men and women.
Answer: Let's define the following events:
\( E_1 \): The selected person is male.
\( E_2 \): The selected person is female.
\( A \): The selected person is an orator.

Since there are equal numbers of men and women:
\( P(E_1) = \frac{1}{2} \)
\( P(E_2) = \frac{1}{2} \)

Now, we find the probability that a person is an orator, given their gender:
\( P(A | E_1) = \text{Probability that a male person is an orator} \)
5 men out of 100 are good orators.
\( P(A | E_1) = \frac{5}{100} = \frac{1}{20} \)

\( P(A | E_2) = \text{Probability that a female person is an orator} \)
25 women out of 1000 are good orators.
\( P(A | E_2) = \frac{25}{1000} = \frac{1}{40} \)

We want to find the probability that the chosen orator is male, which is \( P(E_1 | A) \). Using Bayes' Theorem:
\( P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \)

\( \implies P(E_1 | A) = \frac{\frac{1}{20} \times \frac{1}{2}}{\frac{1}{20} \times \frac{1}{2} + \frac{1}{40} \times \frac{1}{2}} \)

\( \implies P(E_1 | A) = \frac{\frac{1}{40}}{\frac{1}{40} + \frac{1}{80}} \)

\( \implies P(E_1 | A) = \frac{\frac{1}{40}}{\frac{2+1}{80}} \)

\( \implies P(E_1 | A) = \frac{\frac{1}{40}}{\frac{3}{80}} \)

\( \implies P(E_1 | A) = \frac{1}{40} \times \frac{80}{3} \)

\( \implies P(E_1 | A) = \frac{2}{3} \)
In simple words: We used Bayes' Theorem to find the chance that a person is male, given that they are a good speaker. We started by knowing how many men and women are good speakers and that there are equal numbers of men and women. The answer tells us that if an orator is chosen, the chance they are male is 2 out of 3.

🎯 Exam Tip: Remember to simplify fractions for probabilities like \( \frac{5}{100} \) and \( \frac{25}{1000} \) early in your calculations to avoid larger numbers and potential errors.

 

Question 4. Suppose that 5 % of men and 0.25 % of women have grey hair. A grey haired person is selected at random, what is the probability of this person being male? Assume that there are equal number of males and females.
Answer: Let's define the events as follows:
\( E_1 \): The person selected is male.
\( E_2 \): The person selected is female.
\( E \): The person has grey hair.

Since there are equal numbers of males and females:
\( P(E_1) = \frac{1}{2} \)
\( P(E_2) = \frac{1}{2} \)

Now, we find the probability of a person having grey hair, given their gender:
\( P(E | E_1) = \text{Probability that a male person has grey hair} \)
5% of men have grey hair.
\( P(E | E_1) = 5\% = \frac{5}{100} \)

\( P(E | E_2) = \text{Probability that a female person has grey hair} \)
0.25% of women have grey hair.
\( P(E | E_2) = 0.25\% = \frac{0.25}{100} \)

We want to find the probability that a grey-haired person is male, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{\frac{5}{100} \times \frac{1}{2}}{\frac{5}{100} \times \frac{1}{2} + \frac{0.25}{100} \times \frac{1}{2}} \)

\( \implies P(E_1 | E) = \frac{\frac{5}{200}}{\frac{5}{200} + \frac{0.25}{200}} \)

\( \implies P(E_1 | E) = \frac{5}{5 + 0.25} \)

\( \implies P(E_1 | E) = \frac{5}{5.25} \)

To remove the decimal, multiply the numerator and denominator by 100:
\( \implies P(E_1 | E) = \frac{500}{525} \)

Simplify the fraction by dividing by 25:
\( \implies P(E_1 | E) = \frac{20}{21} \)
In simple words: We know the percentage of men and women with grey hair, and that there are equal numbers of men and women. We used Bayes' Theorem to find the probability that a person is male, given that they have grey hair. The calculation shows there's a 20 out of 21 chance that a grey-haired person is male.

🎯 Exam Tip: When dealing with percentages in probability, always convert them to fractions or decimals (e.g., \( 5\% = \frac{5}{100} \)) before performing calculations to avoid errors.

 

Question 5. A company has two plants to manufacture scooters. Plant I manufactures 70 % of the scooters and plant II manufactures 30 %. At plant I, 80 % of scooters are rated standard quality and at plant II, 90 % of scooters are rated standard quality. A scooter is picked up at random and is found to be of standard quality. What is the chance that it was from plant II ?
Answer: Let's define the events as follows:
\( E_1 \): A scooter is manufactured by Plant I.
\( E_2 \): A scooter is manufactured by Plant II.
\( E \): A scooter is of standard quality.

The probabilities of a scooter being manufactured by each plant are:
\( P(E_1) = 70\% = \frac{70}{100} = \frac{7}{10} \)
\( P(E_2) = 30\% = \frac{30}{100} = \frac{3}{10} \)

Now, we find the probability of a scooter being of standard quality from each plant:
\( P(E | E_1) = \text{Probability of standard quality from Plant I} \)
80% of scooters from Plant I are standard quality.
\( P(E | E_1) = 80\% = \frac{80}{100} = \frac{8}{10} \)

\( P(E | E_2) = \text{Probability of standard quality from Plant II} \)
90% of scooters from Plant II are standard quality.
\( P(E | E_2) = 90\% = \frac{90}{100} = \frac{9}{10} \)

We want to find the probability that a standard quality scooter was from Plant II, which is \( P(E_2 | E) \). Using Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_2 | E) = \frac{\frac{9}{10} \times \frac{3}{10}}{\frac{8}{10} \times \frac{7}{10} + \frac{9}{10} \times \frac{3}{10}} \)

\( \implies P(E_2 | E) = \frac{\frac{27}{100}}{\frac{56}{100} + \frac{27}{100}} \)

\( \implies P(E_2 | E) = \frac{\frac{27}{100}}{\frac{56+27}{100}} \)

\( \implies P(E_2 | E) = \frac{\frac{27}{100}}{\frac{83}{100}} \)

\( \implies P(E_2 | E) = \frac{27}{83} \)
In simple words: We used Bayes' Theorem to find the likelihood that a standard quality scooter came from Plant II. We considered how many scooters each plant makes and what percentage of their scooters are good quality. The calculation shows there's a 27 out of 83 chance that the scooter came from Plant II.

🎯 Exam Tip: Keep the denominators consistent (e.g., all as 100 or 10) throughout the fractions to simplify the final calculation. This helps in combining terms in the denominator more easily.

 

Question 6. A company has two plants to manufacture bicycles. The first plant manufactures 60 % of the bicycles and the second plant 40 % .80 % of the bicycles are rated of standard quality and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality. Find the probability that it comes from the second plant.
Answer: Let's define the events as follows:
\( E_1 \): A bicycle is manufactured by the first plant.
\( E_2 \): A bicycle is manufactured by the second plant.
\( A \): A selected bicycle is of standard quality.

The probabilities of a bicycle being manufactured by each plant are:
\( P(E_1) = 60\% = \frac{60}{100} \)
\( P(E_2) = 40\% = \frac{40}{100} \)

Now, we find the probability of a bicycle being of standard quality from each plant:
\( P(A | E_1) = \text{Probability of standard quality from first plant} \)
80% of bicycles from the first plant are standard quality.
\( P(A | E_1) = 80\% = \frac{80}{100} \)

\( P(A | E_2) = \text{Probability of standard quality from second plant} \)
90% of bicycles from the second plant are standard quality.
\( P(A | E_2) = 90\% = \frac{90}{100} \)

We want to find the probability that a standard quality bicycle was from the second plant, which is \( P(E_2 | A) \). Using Bayes' Theorem:
\( P(E_2 | A) = \frac{P(E_2) P(A | E_2)}{P(E_1) P(A | E_1) + P(E_2) P(A | E_2)} \)

\( \implies P(E_2 | A) = \frac{\frac{40}{100} \times \frac{90}{100}}{\frac{60}{100} \times \frac{80}{100} + \frac{40}{100} \times \frac{90}{100}} \)

\( \implies P(E_2 | A) = \frac{\frac{3600}{10000}}{\frac{4800}{10000} + \frac{3600}{10000}} \)

\( \implies P(E_2 | A) = \frac{3600}{4800 + 3600} \)

\( \implies P(E_2 | A) = \frac{3600}{8400} \)

Simplify the fraction by dividing by 100, then by 12:
\( \implies P(E_2 | A) = \frac{36}{84} = \frac{3}{7} \)
In simple words: We used Bayes' Theorem to figure out the chance that a good quality bicycle came from the second factory. We considered how many bikes each factory makes and how many of those are good. The calculation shows there's a 3 out of 7 chance that the good bike was from the second factory.

🎯 Exam Tip: When dealing with percentages, converting them to fractions with a common denominator (like 100) often makes the addition and division steps clearer and reduces calculation errors.

 

Question 7. A has an alarm which will ring at the appointed time with probability 0.9 . If the alarm rings, it will awake him and he will/ each the examination hall in time with probability 0.8 . If the alarm does not ring, A will get up at his own time to reach the examination hall in time, with probability 0.3. Knowing that the person A reached the hall in time, find the probability that the alarm ring.
Answer: Let's define the events as follows:
\( E_1 \): The alarm rings.
\( E_2 \): The alarm does not ring.
\( E \): Person A reached the hall in time.

The probability that the alarm rings is 0.9. So, the probability that it does not ring is:
\( P(E_1) = 0.9 \)
\( P(E_2) = 1 - 0.9 = 0.1 \)

Now, we find the probability of reaching the hall in time, given whether the alarm rang or not:
\( P(E | E_1) = \text{Probability of reaching in time if alarm rings} \)
If the alarm rings, he reaches in time with probability 0.8.
\( P(E | E_1) = 0.8 \)

\( P(E | E_2) = \text{Probability of reaching in time if alarm does not ring} \)
If the alarm does not ring, he reaches in time with probability 0.3.
\( P(E | E_2) = 0.3 \)

We want to find the probability that the alarm rang, given that he reached the hall in time, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{0.8 \times 0.9}{0.8 \times 0.9 + 0.3 \times 0.1} \)

\( \implies P(E_1 | E) = \frac{0.72}{0.72 + 0.03} \)

\( \implies P(E_1 | E) = \frac{0.72}{0.75} \)

To remove the decimal, multiply the numerator and denominator by 100:
\( \implies P(E_1 | E) = \frac{72}{75} \)

Simplify the fraction by dividing by 3:
\( \implies P(E_1 | E) = \frac{24}{25} = 0.96 \)
In simple words: We used Bayes' Theorem to find the chance that the alarm rang, knowing that the person reached the exam hall on time. We considered the probabilities of the alarm ringing and of reaching on time in both situations (alarm rings or not). The calculation shows there's a high chance (24 out of 25) that the alarm rang.

🎯 Exam Tip: Be careful with decimal calculations, especially when adding or dividing. Always write down each step clearly to minimize errors and consider converting to fractions at the end if it helps simplify.

 

Question 8. By examining the chest X-rays, probability that T.B. is detected when a person is actually suffering, is 0.99 . The probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X-ray is 0.001 . In a certain city 1 in 1000 suffers from T.B. A person selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.
Answer: Let's define the events as follows:
\( E_1 \): The person is actually suffering from T.B.
\( E_2 \): The person is not suffering from T.B.
\( A \): The person is diagnosed to have T.B.

The probability that a person suffers from T.B. in the city is:
\( P(E_1) = \frac{1}{1000} \)
The probability that a person does not suffer from T.B. is:
\( P(E_2) = 1 - \frac{1}{1000} = \frac{999}{1000} \)

Now, we find the probability of being diagnosed with T.B. under different conditions:
\( P(A | E_1) = \text{Probability of being diagnosed with T.B. when actually suffering} \)
This is the true positive rate: 0.99.
\( P(A | E_1) = 0.99 \)

\( P(A | E_2) = \text{Probability of being diagnosed with T.B. when not suffering (incorrect diagnosis)} \)
This is the false positive rate: 0.001.
\( P(A | E_2) = 0.001 \)

We want to find the probability that a person actually has T.B., given that they are diagnosed with T.B., which is \( P(E_1 | A) \). Using Bayes' Theorem:
\( P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2)} \)

\( \implies P(E_1 | A) = \frac{0.99 \times \frac{1}{1000}}{0.99 \times \frac{1}{1000} + 0.001 \times \frac{999}{1000}} \)

\( \implies P(E_1 | A) = \frac{\frac{0.99}{1000}}{\frac{0.99}{1000} + \frac{0.999}{1000}} \)

\( \implies P(E_1 | A) = \frac{0.99}{0.99 + 0.999} \)

\( \implies P(E_1 | A) = \frac{0.99}{1.989} \)

To remove the decimal, multiply the numerator and denominator by 1000:
\( \implies P(E_1 | A) = \frac{990}{1989} \)

Simplify the fraction by dividing by 99:
\( \implies P(E_1 | A) = \frac{10}{20.09...} \) (This step does not simplify cleanly if dividing by 99 as it should be by 9 or 33)
Let's recheck the simplification: \( \frac{990}{1989} \). Both are divisible by 9.
\( 990 \div 9 = 110 \)
\( 1989 \div 9 = 221 \)
\( \implies P(E_1 | A) = \frac{110}{221} \)
In simple words: We used Bayes' Theorem to find the chance that someone actually has T.B. if they are diagnosed with it. This involves knowing how often the test is correct, how often it's wrong, and how common T.B. is in the population. The result tells us that even with a positive diagnosis, the chance of actually having T.B. is 110 out of 221.

🎯 Exam Tip: When dealing with very small decimal probabilities, it can be easier to work with fractions or multiply by powers of 10 to convert them to whole numbers for calculations. Pay close attention to the definition of true positive and false positive rates.

 

Question 9. Shoes are produced by two machines A and B .50 % of the shoes are produced by machine A with an estimate of 10 % of them being defective. On machine B, 20 % of the shoes produced are defective if a shoe taken at random is found to be defective, what is the probability that shoe was produced by machine A I
Answer: Let's define the events as follows:
\( E_1 \): Shoes are produced by machine A.
\( E_2 \): Shoes are produced by machine B.
\( E \): A shoe is defective.

The probabilities of a shoe being produced by each machine are:
\( P(E_1) = 50\% = \frac{1}{2} \)
\( P(E_2) = 100\% - 50\% = 50\% = \frac{1}{2} \)

Now, we find the probability of a shoe being defective from each machine:
\( P(E | E_1) = \text{Probability of a defective shoe from machine A} \)
10% of shoes from machine A are defective.
\( P(E | E_1) = 10\% = \frac{10}{100} = \frac{1}{10} \)

\( P(E | E_2) = \text{Probability of a defective shoe from machine B} \)
20% of shoes from machine B are defective.
\( P(E | E_2) = 20\% = \frac{20}{100} = \frac{2}{10} \)

We want to find the probability that a defective shoe was produced by machine A, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{\frac{1}{10} \times \frac{1}{2}}{\frac{1}{10} \times \frac{1}{2} + \frac{2}{10} \times \frac{1}{2}} \)

\( \implies P(E_1 | E) = \frac{\frac{1}{20}}{\frac{1}{20} + \frac{2}{20}} \)

\( \implies P(E_1 | E) = \frac{\frac{1}{20}}{\frac{1+2}{20}} \)

\( \implies P(E_1 | E) = \frac{\frac{1}{20}}{\frac{3}{20}} \)

\( \implies P(E_1 | E) = \frac{1}{3} \)
In simple words: We used Bayes' Theorem to find the chance that a defective shoe came from machine A. We considered how much each machine produces and how many of its shoes are faulty. The calculation shows that if a defective shoe is found, there's a 1 out of 3 chance it came from machine A.

🎯 Exam Tip: Pay attention to the wording "what is the probability that shoe was produced by machine A I", which clearly asks for \( P(E_1 | E) \). Ensure to correctly identify all given percentages as probabilities for each event.

 

Question 10. 15% of the employees are graduates (G), and of these, 80 % work in administrative posts (A). Of the non-graduate (NG) employees of the company, 10 % work in administrative posts. Find the probability that an employee of this company selected at random from those working in administrative posts will be a graduate.
Answer: Let's define the events as follows:
\( E_1 \): The employee is a graduate.
\( E_2 \): The employee is a non-graduate.
\( E \): The employee works in an administrative post.

The probabilities of an employee being a graduate or non-graduate are:
\( P(E_1) = 15\% = \frac{15}{100} \)
\( P(E_2) = 100\% - 15\% = 85\% = \frac{85}{100} \)

Now, we find the probability of working in an administrative post, given their education level:
\( P(E | E_1) = \text{Probability that a graduate works in administrative post} \)
80% of graduates work in administrative posts.
\( P(E | E_1) = 80\% = \frac{80}{100} \)

\( P(E | E_2) = \text{Probability that a non-graduate works in administrative post} \)
10% of non-graduates work in administrative posts.
\( P(E | E_2) = 10\% = \frac{10}{100} \)

We want to find the probability that an employee working in an administrative post is a graduate, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{\frac{80}{100} \times \frac{15}{100}}{\frac{80}{100} \times \frac{15}{100} + \frac{10}{100} \times \frac{85}{100}} \)

\( \implies P(E_1 | E) = \frac{\frac{1200}{10000}}{\frac{1200}{10000} + \frac{850}{10000}} \)

\( \implies P(E_1 | E) = \frac{1200}{1200 + 850} \)

\( \implies P(E_1 | E) = \frac{1200}{2050} \)

Simplify the fraction by dividing by 10, then by 5:
\( \implies P(E_1 | E) = \frac{120}{205} = \frac{24}{41} \)
Alternatively, as a decimal:
\( \implies P(E_1 | E) = 0.585 \)
In simple words: We used Bayes' Theorem to calculate the probability that an employee is a graduate, given they work in an administrative role. We took into account the percentage of graduates and non-graduates, and how many from each group work in administration. The result shows that if you pick an employee from an administrative post, there's about a 58.5% chance they are a graduate.

🎯 Exam Tip: When simplifying fractions, look for common factors like 10 or 5 first. Double-check your arithmetic, especially when adding decimals or large numbers in the denominator.

 

Question 11. Three urns contain 6 red and 4 black, 4 red and 6 black, and 5 red and 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the first urn.
Answer: Let's define the events as follows:
\( E_1 \): Urn I is chosen.
\( E_2 \): Urn II is chosen.
\( E_3 \): Urn III is chosen.
\( E \): A red ball is drawn.

Since one of the three urns is selected at random:
\( P(E_1) = \frac{1}{3} \)
\( P(E_2) = \frac{1}{3} \)
\( P(E_3) = \frac{1}{3} \)

Now, we find the probability of drawing a red ball from each urn:
\( P(E | E_1) = \text{Probability of drawing a red ball from Urn I} \)
Urn I contains 6 red and 4 black balls (total 10 balls).
\( P(E | E_1) = \frac{6}{10} \)

\( P(E | E_2) = \text{Probability of drawing a red ball from Urn II} \)
Urn II contains 4 red and 6 black balls (total 10 balls).
\( P(E | E_2) = \frac{4}{10} \)

\( P(E | E_3) = \text{Probability of drawing a red ball from Urn III} \)
Urn III contains 5 red and 5 black balls (total 10 balls).
\( P(E | E_3) = \frac{5}{10} \)

We want to find the probability that the red ball was drawn from Urn I, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_1 | E) = \frac{\frac{6}{10} \times \frac{1}{3}}{\frac{6}{10} \times \frac{1}{3} + \frac{4}{10} \times \frac{1}{3} + \frac{5}{10} \times \frac{1}{3}} \)

\( \implies P(E_1 | E) = \frac{\frac{6}{30}}{\frac{6}{30} + \frac{4}{30} + \frac{5}{30}} \)

\( \implies P(E_1 | E) = \frac{\frac{6}{30}}{\frac{6+4+5}{30}} \)

\( \implies P(E_1 | E) = \frac{\frac{6}{30}}{\frac{15}{30}} \)

\( \implies P(E_1 | E) = \frac{6}{15} \)

Simplify the fraction by dividing by 3:
\( \implies P(E_1 | E) = \frac{2}{5} \)
In simple words: We used Bayes' Theorem to find the chance that a red ball came from the first urn, knowing that a red ball was picked. We started with the equal chance of picking any of the three urns, and then calculated the chance of drawing a red ball from each. The answer tells us there's a 2 out of 5 chance it came from Urn I.

🎯 Exam Tip: When all initial probabilities \( P(E_i) \) are equal, they can often cancel out in the Bayes' Theorem formula, simplifying the calculation. Ensure all fractions are fully simplified at the end.

 

Question 12. Three mris are given, each containing red and black balls as indicated below: Urn I: 6 rid and 4 black balls. Urn II: 2 red and 6 black balls. Urn III: 1 red and 8 black balls. An urn is chosen at random and a ball is drawn from the urn. The ball drawn is red. Find the probability that the ball is drawn.either from urn II or from urn III.
Answer: Let's define the events as follows:
\( E_1 \): Urn I is chosen.
\( E_2 \): Urn II is chosen.
\( E_3 \): Urn III is chosen.
\( E \): A red ball is drawn.

Since one of the three urns is chosen at random:
\( P(E_1) = \frac{1}{3} \)
\( P(E_2) = \frac{1}{3} \)
\( P(E_3) = \frac{1}{3} \)

Now, we find the probability of drawing a red ball from each urn:
\( P(E | E_1) = \text{Probability of drawing a red ball from Urn I} \)
Urn I contains 6 red and 4 black balls (total 10 balls).
\( P(E | E_1) = \frac{6}{10} \)

\( P(E | E_2) = \text{Probability of drawing a red ball from Urn II} \)
Urn II contains 2 red and 6 black balls (total 8 balls).
\( P(E | E_2) = \frac{2}{8} \)

\( P(E | E_3) = \text{Probability of drawing a red ball from Urn III} \)
Urn III contains 1 red and 8 black balls (total 9 balls).
\( P(E | E_3) = \frac{1}{9} \)

We need to find the probability that the red ball was drawn from either Urn II or Urn III, which is \( P(E_2 | E) + P(E_3 | E) \). First, we calculate each using Bayes' Theorem.

For \( P(E_2 | E) \):
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{8} \times \frac{1}{3}}{\frac{6}{10} \times \frac{1}{3} + \frac{2}{8} \times \frac{1}{3} + \frac{1}{9} \times \frac{1}{3}} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{24}}{\frac{6}{30} + \frac{2}{24} + \frac{1}{27}} \)

\( \implies P(E_2 | E) = \frac{\frac{1}{12}}{\frac{1}{5} + \frac{1}{12} + \frac{1}{27}} \)

To add the fractions in the denominator, find a common multiple for 5, 12, 27. LCM(5, 12, 27) = LCM(5, \( 2^2 \times 3 \), \( 3^3 \)) = \( 5 \times 2^2 \times 3^3 = 5 \times 4 \times 27 = 20 \times 27 = 540 \).
\( \implies P(E_2 | E) = \frac{\frac{1}{12}}{\frac{108}{540} + \frac{45}{540} + \frac{20}{540}} \)

\( \implies P(E_2 | E) = \frac{\frac{1}{12}}{\frac{108+45+20}{540}} \)

\( \implies P(E_2 | E) = \frac{\frac{1}{12}}{\frac{173}{540}} \)

\( \implies P(E_2 | E) = \frac{1}{12} \times \frac{540}{173} \)

\( \implies P(E_2 | E) = \frac{45}{173} \)

For \( P(E_3 | E) \):
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_3 | E) = \frac{\frac{1}{9} \times \frac{1}{3}}{\frac{1}{5} + \frac{1}{12} + \frac{1}{27}} \) (Denominator is the same as above)

\( \implies P(E_3 | E) = \frac{\frac{1}{27}}{\frac{173}{540}} \)

\( \implies P(E_3 | E) = \frac{1}{27} \times \frac{540}{173} \)

\( \implies P(E_3 | E) = \frac{20}{173} \)

Now, we sum these probabilities:
Required probability \( = P(E_2 | E) + P(E_3 | E) = \frac{45}{173} + \frac{20}{173} \)

\( \implies = \frac{45+20}{173} = \frac{65}{173} \)
In simple words: We first found the chance of picking each urn and the chance of getting a red ball from each. Then, using Bayes' Theorem, we figured out the individual chances that the red ball came from Urn II and Urn III. Finally, we added these two chances together to get the total probability that the red ball came from either Urn II or Urn III.

🎯 Exam Tip: When calculating sums of probabilities in the denominator, finding the Least Common Multiple (LCM) of all denominators is crucial for accurate addition of fractions.

 

Question 13. Suppose that there is a chance for a newly constructed house to collapse whether the design is faulty or not. The chance that the design is faulty is 20 %. The chance that the house collapses if the design is faulty is 98 % and otherwise it is 25 %. It is seen that the house collapsed. What is the probability, that it is due to faulty design?
Answer: Let's define the events as follows:
\( E_1 \): The design of the house is faulty.
\( E_2 \): The design of the house is not faulty.
\( E \): The house collapses.

The probabilities related to the design are:
\( P(E_1) = 20\% = 0.2 \)
\( P(E_2) = 1 - 0.2 = 0.8 \)

Now, we find the probability of the house collapsing given the design status:
\( P(E | E_1) = \text{Probability of collapse if design is faulty} \)
98% chance of collapse if the design is faulty.
\( P(E | E_1) = 98\% = 0.98 \)

\( P(E | E_2) = \text{Probability of collapse if design is not faulty} \)
25% chance of collapse if the design is not faulty.
\( P(E | E_2) = 25\% = 0.25 \)

We want to find the probability that the design was faulty, given that the house collapsed, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)

\( \implies P(E_1 | E) = \frac{0.98 \times 0.2}{0.98 \times 0.2 + 0.25 \times 0.8} \)

\( \implies P(E_1 | E) = \frac{0.196}{0.196 + 0.200} \)

\( \implies P(E_1 | E) = \frac{0.196}{0.396} \)

To remove the decimal, multiply the numerator and denominator by 1000:
\( \implies P(E_1 | E) = \frac{196}{396} \)

Simplify the fraction by dividing by 4:
\( \implies P(E_1 | E) = \frac{49}{99} \)
In simple words: We used Bayes' Theorem to find the chance that a house collapsed because of a faulty design. We already knew how likely it is for a design to be faulty and how often houses collapse for both good and bad designs. This calculation helps us see the real chance of a faulty design being the cause after a collapse happens.

🎯 Exam Tip: When working with decimals in Bayes' Theorem, perform multiplication first and then addition/subtraction. Multiplying by a power of 10 to eliminate decimals before the final simplification can prevent errors.

 

Question 14. In an automobile factory, certain parts are to be fixed to the chasis in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45 % B has 35 % and C has 20 % chance of doing it. The probability that A, B and C will take more than the allotted time are 1 / 6, 1 / 10 and 1 / 20 respectively. If it is found that none of them has taken more time, what is the probability that A has taken more time?
Answer: Let's define the events as follows:
\( E_1 \): Person A carries out the task.
\( E_2 \): Person B carries out the task.
\( E_3 \): Person C carries out the task.
\( E \): The task takes more than the allotted time.

The probabilities that each person does the task are:
\( P(E_1) = 45\% = \frac{45}{100} \)
\( P(E_2) = 35\% = \frac{35}{100} \)
\( P(E_3) = 20\% = \frac{20}{100} \)

Now, we find the probability that each person takes more than the allotted time:
\( P(E | E_1) = \text{Probability A takes more time} = \frac{1}{6} \)
\( P(E | E_2) = \text{Probability B takes more time} = \frac{1}{10} \)
\( P(E | E_3) = \text{Probability C takes more time} = \frac{1}{20} \)

We want to find the probability that A has taken more time, given that the task took more time, which is \( P(E_1 | E) \). Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_1 | E) = \frac{\frac{1}{6} \times \frac{45}{100}}{\frac{1}{6} \times \frac{45}{100} + \frac{1}{10} \times \frac{35}{100} + \frac{1}{20} \times \frac{20}{100}} \)

\( \implies P(E_1 | E) = \frac{\frac{45}{600}}{\frac{45}{600} + \frac{35}{1000} + \frac{20}{2000}} \)

Find a common denominator for 600, 1000, 2000. LCM(600, 1000, 2000) = 6000.
\( \implies P(E_1 | E) = \frac{\frac{45}{600}}{\frac{450}{6000} + \frac{210}{6000} + \frac{60}{6000}} \)

\( \implies P(E_1 | E) = \frac{\frac{45}{600}}{\frac{450+210+60}{6000}} \)

\( \implies P(E_1 | E) = \frac{\frac{45}{600}}{\frac{720}{6000}} \)

\( \implies P(E_1 | E) = \frac{45}{600} \times \frac{6000}{720} \)

\( \implies P(E_1 | E) = \frac{45 \times 10}{720} \)

\( \implies P(E_1 | E) = \frac{450}{720} \)

Simplify by dividing by 10, then by 9:
\( \implies P(E_1 | E) = \frac{45}{72} = \frac{5}{8} \)
In simple words: We used Bayes' Theorem to find the probability that person A was working, given that the task took longer than expected. We first listed the chances of each person doing the task and how often each person takes too long. The final answer, 5 out of 8, tells us the likelihood that A was the one who took too long.

🎯 Exam Tip: When fractions have different denominators, find the Least Common Multiple (LCM) to combine them correctly in the denominator of Bayes' formula. Simplify fractions step-by-step to avoid calculation errors.

 

Question 15. Urn A contains 2 white, 1 black and 3 red balls, urn B contains 3 white, 2 black and 4 red balls and urn C contains 4 white, 3 black and 2 red balls. One urn is chosen at random and 2 balls are drawn at random from the urn. If the chosen balls happen to be red and black, what is the probability that both balls come from urn B?
Answer: Let's define the events as follows:
\( E_1 \): Urn A is chosen.
\( E_2 \): Urn B is chosen.
\( E_3 \): Urn C is chosen.
\( E \): Two balls drawn are one red and one black.

Since one of the three urns is chosen at random:
\( P(E_1) = \frac{1}{3} \)
\( P(E_2) = \frac{1}{3} \)
\( P(E_3) = \frac{1}{3} \)

Now, we calculate the probability of drawing one red and one black ball from each urn.
Total balls in Urn A: \( 2(\text{white}) + 1(\text{black}) + 3(\text{red}) = 6 \) balls.
\( P(E | E_1) = \text{Probability of drawing 1 red and 1 black from Urn A} \)
We choose 1 red out of 3, and 1 black out of 1. Total ways to choose 2 from 6 is \( ^{6}C_2 \).
\( P(E | E_1) = \frac{^{3}C_1 \times ^{1}C_1}{^{6}C_2} = \frac{3 \times 1}{\frac{6 \times 5}{2}} = \frac{3}{15} = \frac{1}{5} \)

Total balls in Urn B: \( 3(\text{white}) + 2(\text{black}) + 4(\text{red}) = 9 \) balls.
\( P(E | E_2) = \text{Probability of drawing 1 red and 1 black from Urn B} \)
We choose 1 red out of 4, and 1 black out of 2. Total ways to choose 2 from 9 is \( ^{9}C_2 \).
\( P(E | E_2) = \frac{^{4}C_1 \times ^{2}C_1}{^{9}C_2} = \frac{4 \times 2}{\frac{9 \times 8}{2}} = \frac{8}{36} = \frac{2}{9} \)

Total balls in Urn C: \( 4(\text{white}) + 3(\text{black}) + 2(\text{red}) = 9 \) balls.
\( P(E | E_3) = \text{Probability of drawing 1 red and 1 black from Urn C} \)
We choose 1 red out of 2, and 1 black out of 3. Total ways to choose 2 from 9 is \( ^{9}C_2 \).
\( P(E | E_3) = \frac{^{2}C_1 \times ^{3}C_1}{^{9}C_2} = \frac{2 \times 3}{\frac{9 \times 8}{2}} = \frac{6}{36} = \frac{1}{6} \)

We want to find the probability that the balls came from Urn B, given that two balls drawn are one red and one black, which is \( P(E_2 | E) \). Using Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{2}{9} \times \frac{1}{3} + \frac{1}{6} \times \frac{1}{3}} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{27}}{\frac{1}{15} + \frac{2}{27} + \frac{1}{18}} \)

Find a common denominator for 15, 27, 18. LCM(15, 27, 18) = LCM(\( 3 \times 5 \), \( 3^3 \), \( 2 \times 3^2 \)) = \( 2 \times 3^3 \times 5 = 2 \times 27 \times 5 = 270 \).
\( \implies P(E_2 | E) = \frac{\frac{2}{27}}{\frac{18}{270} + \frac{20}{270} + \frac{15}{270}} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{27}}{\frac{18+20+15}{270}} \)

\( \implies P(E_2 | E) = \frac{\frac{2}{27}}{\frac{53}{270}} \)

\( \implies P(E_2 | E) = \frac{2}{27} \times \frac{270}{53} \)

\( \implies P(E_2 | E) = \frac{2 \times 10}{53} \)

\( \implies P(E_2 | E) = \frac{20}{53} \)
In simple words: We want to find the chance that two specific balls (one red, one black) came from Urn B, knowing that we already drew those balls. We first found the probability of picking each urn, and then the probability of drawing one red and one black ball from each. Using Bayes' Theorem, we combined these to get the final answer.

🎯 Exam Tip: Remember to use combinations (\( ^{n}C_r \)) correctly when calculating the probabilities of drawing multiple balls. Clearly showing each step of fraction addition and multiplication helps prevent errors.

 

Question 16. Three bags contain balis as shown in the following table :

A bag is chosen at random and two balts are drawn. They happen to be white and red. What is the probability that they came from the third bag?
Answer: Let's define the events as follows:
\( E_1 \): Bag I is chosen.
\( E_2 \): Bag II is chosen.
\( E_3 \): Bag III is chosen.
\( E \): Two drawn balls are one white and one red.

Since a bag is chosen at random from three bags:
\( P(E_1) = \frac{1}{3} \)
\( P(E_2) = \frac{1}{3} \)
\( P(E_3) = \frac{1}{3} \)

Now, we calculate the probability of drawing one white and one red ball from each bag.
Total balls in Bag I: \( 1(\text{white}) + 2(\text{black}) + 3(\text{red}) = 6 \) balls.
\( P(E | E_1) = \text{Probability of drawing 1 white and 1 red from Bag I} \)
We choose 1 white out of 1, and 1 red out of 3. Total ways to choose 2 from 6 is \( ^{6}C_2 \).
\( P(E | E_1) = \frac{^{1}C_1 \times ^{3}C_1}{^{6}C_2} = \frac{1 \times 3}{\frac{6 \times 5}{2}} = \frac{3}{15} = \frac{1}{5} \)

Total balls in Bag II: \( 2(\text{white}) + 1(\text{black}) + 1(\text{red}) = 4 \) balls.
\( P(E | E_2) = \text{Probability of drawing 1 white and 1 red from Bag II} \)
We choose 1 white out of 2, and 1 red out of 1. Total ways to choose 2 from 4 is \( ^{4}C_2 \).
\( P(E | E_2) = \frac{^{2}C_1 \times ^{1}C_1}{^{4}C_2} = \frac{2 \times 1}{\frac{4 \times 3}{2}} = \frac{2}{6} = \frac{1}{3} \)

Total balls in Bag III: \( 4(\text{white}) + 3(\text{black}) + 2(\text{red}) = 9 \) balls.
\( P(E | E_3) = \text{Probability of drawing 1 white and 1 red from Bag III} \)
We choose 1 white out of 4, and 1 red out of 2. Total ways to choose 2 from 9 is \( ^{9}C_2 \).
\( P(E | E_3) = \frac{^{4}C_1 \times ^{2}C_1}{^{9}C_2} = \frac{4 \times 2}{\frac{9 \times 8}{2}} = \frac{8}{36} = \frac{2}{9} \)

We want to find the probability that the balls came from Bag III, given that two balls drawn are one white and one red, which is \( P(E_3 | E) \). Using Bayes' Theorem:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)

\( \implies P(E_3 | E) = \frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{3} + \frac{2}{9} \times \frac{1}{3}} \)

\( \implies P(E_3 | E) = \frac{\frac{2}{27}}{\frac{1}{15} + \frac{1}{9} + \frac{2}{27}} \)

Find a common denominator for 15, 9, 27. LCM(15, 9, 27) = LCM(\( 3 \times 5 \), \( 3^2 \), \( 3^3 \)) = \( 3^3 \times 5 = 27 \times 5 = 135 \).
\( \implies P(E_3 | E) = \frac{\frac{2}{27}}{\frac{9}{135} + \frac{15}{135} + \frac{10}{135}} \)

\( \implies P(E_3 | E) = \frac{\frac{2}{27}}{\frac{9+15+10}{135}} \)

\( \implies P(E_3 | E) = \frac{\frac{2}{27}}{\frac{34}{135}} \)

\( \implies P(E_3 | E) = \frac{2}{27} \times \frac{135}{34} \)

\( \implies P(E_3 | E) = \frac{2 \times 5}{34} \)

\( \implies P(E_3 | E) = \frac{10}{34} \)

Simplify by dividing by 2:
\( \implies P(E_3 | E) = \frac{5}{17} \)
In simple words: We used Bayes' Theorem to find the chance that the two balls (one white and one red) we drew came from the third bag. We first looked at how likely it was to choose each bag and then how likely it was to draw those specific balls from each bag. The result tells us the probability of the balls being from the third bag.

🎯 Exam Tip: Remember that "balis" in the question text is a typo for "balls". Clearly state the total number of balls in each bag before calculating combinations, as this is a common source of error. Always simplify fractions to their lowest terms.

 

Question 15. Urn A contains 2 white, 1 black and 3 red balls, urn B contains 3 white, 2 black and 4 red balls and urn C contains 4 white, 3 black and 2 red balls. One urn is chosen at random and 2 balls are drawn at random from the urn. If the chosen balls happen to be red and black, what is the probability that both balls come from urn B?
Answer:
Let's define the following events:
\( E_1 \): Urn A is chosen.
\( E_2 \): Urn B is chosen.
\( E_3 \): Urn C is chosen.
\( E \): Two balls drawn are one red and one black.
Since one urn is chosen at random from three, the probability for each urn is \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
For Urn A (2 white, 1 black, 3 red = 6 balls total):
The probability of drawing one red and one black ball from Urn A is \( P(E | E_1) = \frac{{ }^3 C_1 \times { }^1 C_1}{{ }^6 C_2} \).
\( \implies P(E | E_1) = \frac{3 \times 1}{\frac{6 \times 5}{2}} = \frac{3}{15} = \frac{1}{5} \)
For Urn B (3 white, 2 black, 4 red = 9 balls total):
The probability of drawing one red and one black ball from Urn B is \( P(E | E_2) = \frac{{ }^4 C_1 \times { }^2 C_1}{{ }^9 C_2} \).
\( \implies P(E | E_2) = \frac{4 \times 2}{\frac{9 \times 8}{2}} = \frac{8}{36} = \frac{2}{9} \)
For Urn C (4 white, 3 black, 2 red = 9 balls total):
The probability of drawing one red and one black ball from Urn C is \( P(E | E_3) = \frac{{ }^2 C_1 \times { }^3 C_1}{{ }^9 C_2} \).
\( \implies P(E | E_3) = \frac{2 \times 3}{\frac{9 \times 8}{2}} = \frac{6}{36} = \frac{1}{6} \)
We want to find the probability that the balls came from Urn B, given that they are one red and one black. This is \( P(E_2 | E) \).
Using Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_2 | E) = \frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{2}{9} \times \frac{1}{3} + \frac{1}{6} \times \frac{1}{3}} \)
We can cancel the \( \frac{1}{3} \) from numerator and denominator:
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{1}{5} + \frac{2}{9} + \frac{1}{6}} \)
To sum the fractions in the denominator, find a common denominator, which is 90:
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{18}{90} + \frac{20}{90} + \frac{15}{90}} \)
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{18+20+15}{90}} = \frac{\frac{2}{9}}{\frac{53}{90}} \)
\( \implies P(E_2 | E) = \frac{2}{9} \times \frac{90}{53} = \frac{2 \times 10}{53} = \frac{20}{53} \)
In simple words: We calculated the chance of drawing one red and one black ball from each urn. Then, using Bayes' Theorem, we found the probability that the chosen balls came from Urn B, given that they were one red and one black. This helps us narrow down the source of the drawn balls.

🎯 Exam Tip: When dealing with combinations and conditional probability, clearly define your events and use the combinations formula \( { }^n C_r = \frac{n!}{r!(n-r)!} \) accurately. Remember to simplify fractions at each step to avoid errors in complex calculations.

 

Question 16. Three bags contain balls as shown in the following table :

A bag is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they came from the third bag?
Answer:
Let's define the events as follows:
\( E_1 \): Bag I is chosen.
\( E_2 \): Bag II is chosen.
\( E_3 \): Bag III is chosen.
\( E \): Two drawn balls are one white and one red.
Since a bag is chosen at random, the probability for each bag is \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
For Bag I (1 white, 2 black, 3 red = 6 balls total):
The probability of drawing one white and one red ball from Bag I is \( P(E | E_1) = \frac{{ }^1 C_1 \times { }^3 C_1}{{ }^6 C_2} \).
\( \implies P(E | E_1) = \frac{1 \times 3}{\frac{6 \times 5}{2}} = \frac{3}{15} = \frac{1}{5} \)
For Bag II (2 white, 1 black, 1 red = 4 balls total):
The probability of drawing one white and one red ball from Bag II is \( P(E | E_2) = \frac{{ }^2 C_1 \times { }^1 C_1}{{ }^4 C_2} \).
\( \implies P(E | E_2) = \frac{2 \times 1}{\frac{4 \times 3}{2}} = \frac{2}{6} = \frac{1}{3} \)
For Bag III (4 white, 3 black, 2 red = 9 balls total):
The probability of drawing one white and one red ball from Bag III is \( P(E | E_3) = \frac{{ }^4 C_1 \times { }^2 C_1}{{ }^9 C_2} \).
\( \implies P(E | E_3) = \frac{4 \times 2}{\frac{9 \times 8}{2}} = \frac{8}{36} = \frac{2}{9} \)
We want to find the probability that the balls came from the third bag, given that they are one white and one red. This is \( P(E_3 | E) \).
Using Bayes' Theorem:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_3 | E) = \frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{3} + \frac{2}{9} \times \frac{1}{3}} \)
We can cancel the \( \frac{1}{3} \) from numerator and denominator:
\( \implies P(E_3 | E) = \frac{\frac{2}{9}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{9}} \)
To sum the fractions in the denominator, find a common denominator, which is 45:
\( \implies P(E_3 | E) = \frac{\frac{2}{9}}{\frac{9}{45} + \frac{15}{45} + \frac{10}{45}} \)
\( \implies P(E_3 | E) = \frac{\frac{2}{9}}{\frac{9+15+10}{45}} = \frac{\frac{2}{9}}{\frac{34}{45}} \)
\( \implies P(E_3 | E) = \frac{2}{9} \times \frac{45}{34} = \frac{2 \times 5}{34} = \frac{10}{34} = \frac{5}{17} \)
In simple words: First, we figured out the chance of picking one white and one red ball from each bag. Then, using Bayes' Theorem, we found the likelihood that the chosen balls came from Bag III, given that they were indeed one white and one red. This helps to identify the original source of the balls.

🎯 Exam Tip: When problems involve selecting items from different sources (like bags or urns) and then calculating a conditional probability, always use Bayes' Theorem. Remember to define your events clearly and carefully calculate the probabilities for each stage.

 

Question 17. In a bolt factory, machines A, B and C manufacture 25 %, 35 % and 40 % respectively. Of the total of their output 5 %, 4 %, and 2 % are defective. A bolt is drawn and found to be defective.
(i) What are the probabilities that it was manufactured by the machines A, B and C ?
(ii) Find the probability that it was manufactured by either machine A or C.
Answer:
(i) Let's define the events:
\( E_1 \): Bolt manufactured by machine A.
\( E_2 \): Bolt manufactured by machine B.
\( E_3 \): Bolt manufactured by machine C.
\( E \): Bolt is found to be defective.
Given probabilities of production by each machine:
\( P(E_1) = 25 \% = \frac{25}{100} \)
\( P(E_2) = 35 \% = \frac{35}{100} \)
\( P(E_3) = 40 \% = \frac{40}{100} \)
Given probabilities of defective bolts from each machine:
\( P(E | E_1) = 5 \% = \frac{5}{100} \)
\( P(E | E_2) = 4 \% = \frac{4}{100} \)
\( P(E | E_3) = 2 \% = \frac{2}{100} \)
We want to find \( P(E_1 | E) \), \( P(E_2 | E) \), and \( P(E_3 | E) \). These are the probabilities that the defective bolt came from machines A, B, and C, respectively.
Using Bayes' Theorem:
First, calculate the total probability of drawing a defective bolt, \( P(E) = P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3) \).
\( P(E) = \frac{5}{100} \times \frac{25}{100} + \frac{4}{100} \times \frac{35}{100} + \frac{2}{100} \times \frac{40}{100} \)
\( \implies P(E) = \frac{125}{10000} + \frac{140}{10000} + \frac{80}{10000} = \frac{125+140+80}{10000} = \frac{345}{10000} \)

Probability that the defective bolt came from machine A:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E)} \)
\( \implies P(E_1 | E) = \frac{\frac{5}{100} \times \frac{25}{100}}{\frac{345}{10000}} = \frac{\frac{125}{10000}}{\frac{345}{10000}} = \frac{125}{345} = \frac{25}{69} \)

Probability that the defective bolt came from machine B:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E)} \)
\( \implies P(E_2 | E) = \frac{\frac{4}{100} \times \frac{35}{100}}{\frac{345}{10000}} = \frac{\frac{140}{10000}}{\frac{345}{10000}} = \frac{140}{345} = \frac{28}{69} \)

Probability that the defective bolt came from machine C:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E)} \)
\( \implies P(E_3 | E) = \frac{\frac{2}{100} \times \frac{40}{100}}{\frac{345}{10000}} = \frac{\frac{80}{10000}}{\frac{345}{10000}} = \frac{80}{345} = \frac{16}{69} \)

(ii) We need to find the probability that the defective bolt was manufactured by either machine A or C. This is \( P(E_1 | E) + P(E_3 | E) \).
\( \text{Required Probability} = \frac{25}{69} + \frac{16}{69} = \frac{25+16}{69} = \frac{41}{69} \)
In simple words: We first found the chance that a defective bolt came from each machine. Then, to find the chance it came from either machine A or C, we added up their individual probabilities because these events cannot happen at the same time. This process helps identify the main source of defective items.

🎯 Exam Tip: For problems with multiple conditional probabilities, first calculate the overall probability of the event (like drawing a defective bolt). Then, apply Bayes' Theorem for each specific condition. For "either/or" scenarios, remember to sum the individual probabilities if the events are mutually exclusive.

 

Question 18. A company has two plants to manufacture cars. Plant I manufactures 80 per cent of the cars and plant II manufactures 20 per cent. At plant I, 85 out of 100 cars are rated standard quality or better. At plant II, only 65 out of 100 cars are rated standard or better.
(i) What is the probability that cars selected at random came from plant I and it is known that the car is of standard quality?
(ii) What is the probability that the cars came from plant II if it is known that the car is of standard quality?
Answer:
Let's define the events:
\( E_1 \): Car manufactured by Plant I.
\( E_2 \): Car manufactured by Plant II.
\( E \): Car is of standard quality or better.
Given probabilities of car manufacturing by each plant:
\( P(E_1) = 80 \% = \frac{80}{100} \)
\( P(E_2) = 20 \% = \frac{20}{100} \)
Given probabilities of standard quality cars from each plant:
\( P(E | E_1) = 85 \% = \frac{85}{100} \)
\( P(E | E_2) = 65 \% = \frac{65}{100} \)

(i) We want to find the probability that a standard quality car came from Plant I. This is \( P(E_1 | E) \).
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{\frac{85}{100} \times \frac{80}{100}}{\frac{85}{100} \times \frac{80}{100} + \frac{65}{100} \times \frac{20}{100}} \)
\( \implies P(E_1 | E) = \frac{\frac{6800}{10000}}{\frac{6800}{10000} + \frac{1300}{10000}} = \frac{6800}{6800+1300} = \frac{6800}{8100} = \frac{68}{81} \)

(ii) We want to find the probability that a standard quality car came from Plant II. This is \( P(E_2 | E) \).
Using Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_2 | E) = \frac{\frac{65}{100} \times \frac{20}{100}}{\frac{85}{100} \times \frac{80}{100} + \frac{65}{100} \times \frac{20}{100}} \)
\( \implies P(E_2 | E) = \frac{\frac{1300}{10000}}{\frac{6800}{10000} + \frac{1300}{10000}} = \frac{1300}{6800+1300} = \frac{1300}{8100} = \frac{13}{81} \)
In simple words: We calculated the chance that a car came from each plant, given it was of standard quality. These kinds of calculations help companies understand which parts of their production process are most effective.

🎯 Exam Tip: Clearly distinguish between \( P(E_i) \) (prior probability of an event) and \( P(E | E_i) \) (conditional probability of an outcome given an event). This helps in correctly setting up the numerator and denominator for Bayes' Theorem. Ensure all percentages are converted to decimals or fractions.

 

Question 19. A factory has three machines X, Y, and Z producing 1000, 2000 and 3000 bolts per day respectively. The machines X produces 1 % defective bolts, Y produces 1.5 % and Z produces 2 % defective bolts. At the end of a day, a bolt is drawn at random and is found defective. What is the probability that this defective bolt has been produced by the machine X?
Answer:
Let's define the events:
\( E_1 \): Bolt produced by machine X.
\( E_2 \): Bolt produced by machine Y.
\( E_3 \): Bolt produced by machine Z.
\( A \): A bolt drawn is defective.
Total bolts produced per day = \( 1000 + 2000 + 3000 = 6000 \).
Probabilities of bolts being produced by each machine:
\( P(E_1) = \frac{1000}{6000} = \frac{1}{6} \)
\( P(E_2) = \frac{2000}{6000} = \frac{1}{3} \)
\( P(E_3) = \frac{3000}{6000} = \frac{1}{2} \)
Probabilities of a bolt being defective from each machine:
\( P(A | E_1) = 1 \% = \frac{1}{100} \)
\( P(A | E_2) = 1.5 \% = \frac{1.5}{100} = \frac{3}{200} \)
\( P(A | E_3) = 2 \% = \frac{2}{100} \)
We want to find the probability that the defective bolt was produced by machine X. This is \( P(E_1 | A) \).
Using Bayes' Theorem:
\( P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3)} \)
\( \implies P(E_1 | A) = \frac{\frac{1}{100} \times \frac{1}{6}}{\frac{1}{100} \times \frac{1}{6} + \frac{1.5}{100} \times \frac{1}{3} + \frac{2}{100} \times \frac{1}{2}} \)
\( \implies P(E_1 | A) = \frac{\frac{1}{600}}{\frac{1}{600} + \frac{1.5}{300} + \frac{1}{100}} \)
Convert all denominators to 600:
\( \implies P(E_1 | A) = \frac{\frac{1}{600}}{\frac{1}{600} + \frac{3}{600} + \frac{6}{600}} \)
\( \implies P(E_1 | A) = \frac{\frac{1}{600}}{\frac{1+3+6}{600}} = \frac{\frac{1}{600}}{\frac{10}{600}} = \frac{1}{10} \)
In simple words: We figured out how likely each machine is to produce a bolt and how often each machine makes a defective bolt. Then, using Bayes' Theorem, we found the chance that a defective bolt came specifically from machine X. This helps a factory identify where most defects are coming from.

🎯 Exam Tip: When calculating probabilities with multiple production sources, ensure all proportions (production shares, defect rates) are correctly converted to probabilities before applying Bayes' Theorem. Simplify fractions to make calculations easier.

 

Question 20. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer:
Let's define the events:
\( E_1 \): The lost card is a diamond.
\( E_2 \): The lost card is not a diamond.
\( E \): Two cards drawn from the remaining pack are both diamonds.
Initially, there are 13 diamonds and 39 non-diamonds in a pack of 52 cards.
Probability that the lost card is a diamond:
\( P(E_1) = \frac{13}{52} = \frac{1}{4} \)
Probability that the lost card is not a diamond:
\( P(E_2) = 1 - P(E_1) = 1 - \frac{1}{4} = \frac{3}{4} \)

Now, consider the conditional probabilities of event E:
If \( E_1 \) (lost card is a diamond) occurred: The remaining pack has 51 cards, with 12 diamonds and 39 non-diamonds.
\( P(E | E_1) = \text{Probability of drawing 2 diamonds from 12 diamonds in 51 cards} = \frac{{}^{12} C_2}{{}^{51} C_2} \)
\( {}^{12} C_2 = \frac{12 \times 11}{2} = 66 \)
\( {}^{51} C_2 = \frac{51 \times 50}{2} = 1275 \)
\( \implies P(E | E_1) = \frac{66}{1275} \)

If \( E_2 \) (lost card is not a diamond) occurred: The remaining pack has 51 cards, with 13 diamonds and 38 non-diamonds.
\( P(E | E_2) = \text{Probability of drawing 2 diamonds from 13 diamonds in 51 cards} = \frac{{}^{13} C_2}{{}^{51} C_2} \)
\( {}^{13} C_2 = \frac{13 \times 12}{2} = 78 \)
\( {}^{51} C_2 = 1275 \)
\( \implies P(E | E_2) = \frac{78}{1275} \)

We want to find \( P(E_1 | E) \), the probability that the lost card was a diamond, given that two drawn cards are diamonds.
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{\frac{66}{1275} \times \frac{1}{4}}{\frac{66}{1275} \times \frac{1}{4} + \frac{78}{1275} \times \frac{3}{4}} \)
We can cancel \( \frac{1}{1275 \times 4} \) from numerator and denominator:
\( \implies P(E_1 | E) = \frac{66}{66 + 78 \times 3} = \frac{66}{66 + 234} = \frac{66}{300} \)
\( \implies P(E_1 | E) = \frac{11}{50} \)
In simple words: We calculated the chance that the lost card was a diamond, given that two diamonds were drawn from the rest of the deck. This is a classic example of how new information (drawing two diamonds) changes our belief about a past event (the identity of the lost card).

🎯 Exam Tip: When a problem involves an item being 'lost' or 'removed', always adjust the total number of items and the number of specific items (like diamonds) for the subsequent probability calculations. Combinations are crucial for counting ways to draw multiple items simultaneously.

 

Question 21. A speaks the truth 2 out of 3 times and B, 4 out of 5 times. They agree in the assertion that from a bag containing 6 balls of different colours a black ball has been drawn. Find the probability that the statements are true.
Answer:
Let's define the events:
\( E \): A black ball is drawn.
\( E' \): A non-black ball is drawn.
From a bag containing 6 balls of different colors, if one is black:
\( P(E) = \text{Probability of drawing a black ball} = \frac{1}{6} \).
\( P(E') = \text{Probability of drawing a non-black ball} = 1 - \frac{1}{6} = \frac{5}{6} \).

Let \( T_A \) be the event that A speaks the truth, and \( T_B \) be the event that B speaks the truth.
\( P(T_A) = \frac{2}{3} \)
\( P(T_B) = \frac{4}{5} \)
\( P(\text{A lies}) = 1 - \frac{2}{3} = \frac{1}{3} \)
\( P(\text{B lies}) = 1 - \frac{4}{5} = \frac{1}{5} \)

Case 1: Both A and B state that a black ball was drawn, and this statement is TRUE. (Let this be event \( I \)).
This means a black ball was actually drawn, AND A told the truth, AND B told the truth.
\( P(I) = P(E) \times P(T_A) \times P(T_B) \)
\( P(I) = \frac{1}{6} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{90} = \frac{4}{45} \)

Case 2: Both A and B state that a black ball was drawn, but this statement is FALSE. (Let this be event \( II \)).
This means a non-black ball was actually drawn, AND A lied, AND B lied.
\( P(II) = P(E') \times P(\text{A lies}) \times P(\text{B lies}) \)
\( P(II) = \frac{5}{6} \times \frac{1}{3} \times \frac{1}{5} = \frac{5}{90} = \frac{1}{18} \)

We want to find the probability that their statement is true, given that they both agree. This is \( P(I | \text{they agree}) \).
The event "they agree" means either Case 1 (statement true) or Case 2 (statement false, but they both lied in agreement).
So, \( P(\text{they agree}) = P(I) + P(II) \).
\( P(\text{they agree}) = \frac{4}{45} + \frac{1}{18} \)
To sum the fractions, find a common denominator, which is 90:
\( P(\text{they agree}) = \frac{8}{90} + \frac{5}{90} = \frac{13}{90} \)

Now, apply the conditional probability formula:
\( P(\text{statement is true | they agree}) = \frac{P(I)}{P(I) + P(II)} \)
\( \implies P(\text{statement is true | they agree}) = \frac{\frac{4}{45}}{\frac{13}{90}} \)
\( \implies P(\text{statement is true | they agree}) = \frac{4}{45} \times \frac{90}{13} = \frac{4 \times 2}{13} = \frac{8}{13} \)
In simple words: We looked at two possibilities: both A and B telling the truth about drawing a black ball, or both A and B lying about it. Since they both said a black ball was drawn, we calculated the chance that their shared statement was actually true. This helps us decide how much to trust their combined report.

🎯 Exam Tip: In truth-teller problems, always consider two scenarios: (1) the event occurred and they spoke truth, and (2) the event did not occur and they lied. Both contribute to them making the same statement. Bayes' Theorem helps combine these to find the true probability.

 

Question 22. In 2004, there will be three candidates for the position of principal C1, C2 and C3. The chances of their selection are in the proportion 4 : 2 : 3 respectively. The probability that C1, if selected, will introduce co-education in the college is 0.3 . The probabilities of C2 and C3 doing the same are respectively 0.5 and 0.8 .
(i) What is the probability that there will be co-education in the college in 2004?
(ii) Also, find the probability that principal C2 introduces co-education in the college.
Answer:
Let's define the events:
\( E_1 \): Candidate C1 is selected as principal.
\( E_2 \): Candidate C2 is selected as principal.
\( E_3 \): Candidate C3 is selected as principal.
\( E \): Co-education is introduced in the college.
The chances of selection are in the proportion 4 : 2 : 3. The total parts are \( 4+2+3 = 9 \).
So, the probabilities of selection are:
\( P(E_1) = \frac{4}{9} \)
\( P(E_2) = \frac{2}{9} \)
\( P(E_3) = \frac{3}{9} = \frac{1}{3} \)
Probabilities of introducing co-education given each candidate is selected:
\( P(E | E_1) = 0.3 = \frac{3}{10} \)
\( P(E | E_2) = 0.5 = \frac{5}{10} \)
\( P(E | E_3) = 0.8 = \frac{8}{10} \)

(i) We want to find the total probability that co-education will be introduced, \( P(E) \).
Using the law of total probability:
\( P(E) = P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3) \)
\( \implies P(E) = \left(\frac{3}{10} \times \frac{4}{9}\right) + \left(\frac{5}{10} \times \frac{2}{9}\right) + \left(\frac{8}{10} \times \frac{3}{9}\right) \)
\( \implies P(E) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{12+10+24}{90} = \frac{46}{90} = \frac{23}{45} \)

(ii) We want to find the probability that Principal C2 introduced co-education, given that co-education was introduced. This is \( P(E_2 | E) \).
Using Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E)} \)
\( \implies P(E_2 | E) = \frac{\frac{5}{10} \times \frac{2}{9}}{\frac{23}{45}} = \frac{\frac{10}{90}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} \)
\( \implies P(E_2 | E) = \frac{1}{9} \times \frac{45}{23} = \frac{5}{23} \)
In simple words: We first calculated the overall chance that co-education would start, by considering each candidate's likelihood of being chosen and their plans. Then, we found the specific chance that C2 was the principal who started it, given that co-education did happen. This shows how we can work backward from an outcome to find the most likely cause.

🎯 Exam Tip: When given proportions for probabilities, always convert them to fractions (or decimals) that sum to 1 before using them in calculations. For conditional probabilities like part (ii), ensure the total probability from part (i) is correctly used in the denominator of Bayes' Theorem.

 

Question 23. You note that your officer is happy on 60 % of your calls, so you assign a probability of his being happy on your visit as 0.6 or 6 / 10. You have noticed also that if he is happy, he accedes to your request with a probability of 0.4 or 4 / 10 whereas if he is not happy, he accedes to the request with a probability of 0.1 or 1/10. You call one day, and he accedes to your request. What is the probability that his being happy?
Answer:
Let's define the events:
\( E_1 \): The officer is happy.
\( E_2 \): The officer is not happy.
\( E \): The officer accedes to the request.
Given probabilities:
\( P(E_1) = 0.6 = \frac{6}{10} \)
\( P(E_2) = 1 - P(E_1) = 1 - 0.6 = 0.4 = \frac{4}{10} \)
Probabilities of acceding to the request based on happiness:
\( P(E | E_1) = 0.4 = \frac{4}{10} \) (If happy, he accedes)
\( P(E | E_2) = 0.1 = \frac{1}{10} \) (If not happy, he accedes)
We want to find the probability that the officer was happy, given that he acceded to the request. This is \( P(E_1 | E) \).
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{\frac{4}{10} \times \frac{6}{10}}{\left(\frac{4}{10} \times \frac{6}{10}\right) + \left(\frac{1}{10} \times \frac{4}{10}\right)} \)
\( \implies P(E_1 | E) = \frac{\frac{24}{100}}{\frac{24}{100} + \frac{4}{100}} = \frac{\frac{24}{100}}{\frac{28}{100}} \)
\( \implies P(E_1 | E) = \frac{24}{28} = \frac{6}{7} \)
In simple words: We know how often the officer is happy and how likely he is to agree when happy or unhappy. When he agrees, we use this information to figure out the chance that he was happy at that moment. This helps in understanding situations where an outcome can result from different initial conditions.

🎯 Exam Tip: Carefully define complementary events (like 'happy' and 'not happy') and their probabilities. Ensure the conditional probabilities are assigned to the correct conditions in Bayes' Theorem. A common mistake is to swap \( P(E | E_i) \) with \( P(E_i | E) \).

 

Question 24. The chance that a female worker in a chemical factory will contract an occupational disease is 0.04 and the chance for a male worker 0.06 . Out of 1000 workers in a factory 200 are females. One worker is selected at random and the worker is found to have contracted the disease. What is the probability that the worker is a female?
Answer:
Let's define the events:
\( E_1 \): The selected worker is female.
\( E_2 \): The selected worker is male.
\( E \): The worker has contracted the disease.
Total workers = 1000. Females = 200. Males = \( 1000 - 200 = 800 \).
Probabilities of selecting a worker type:
\( P(E_1) = \frac{200}{1000} = \frac{2}{10} \)
\( P(E_2) = \frac{800}{1000} = \frac{8}{10} \)
Probabilities of contracting the disease given worker type:
\( P(E | E_1) = 0.04 \) (Female contracts disease)
\( P(E | E_2) = 0.06 \) (Male contracts disease)
We want to find the probability that the worker is female, given that they have contracted the disease. This is \( P(E_1 | E) \).
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{0.04 \times \frac{2}{10}}{\left(0.04 \times \frac{2}{10}\right) + \left(0.06 \times \frac{8}{10}\right)} \)
\( \implies P(E_1 | E) = \frac{\frac{0.08}{10}}{\frac{0.08}{10} + \frac{0.48}{10}} = \frac{0.08}{0.08 + 0.48} \)
\( \implies P(E_1 | E) = \frac{0.08}{0.56} = \frac{8}{56} = \frac{1}{7} \)
In simple words: We calculated the chance that a worker who got sick is a female. We started by knowing how many males and females work there and their individual chances of getting sick. Then, we used that information to find the specific gender of a sick worker.

🎯 Exam Tip: Be careful to correctly calculate the initial probabilities of selecting a male or female worker based on their numbers in the total workforce. When dealing with decimals in fractions, ensure consistent place values or convert to whole numbers to prevent calculation errors.

 

Question 25. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter driver, car driver and a truck driver is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
Let's define the events:
\( E_1 \): The insured person is a scooter driver.
\( E_2 \): The insured person is a car driver.
\( E_3 \): The insured person is a truck driver.
\( E \): The insured person meets with an accident.
Total insured persons = \( 2000 + 4000 + 6000 = 12000 \).
Probabilities of selecting each driver type:
\( P(E_1) = \frac{2000}{12000} = \frac{1}{6} \)
\( P(E_2) = \frac{4000}{12000} = \frac{1}{3} \)
\( P(E_3) = \frac{6000}{12000} = \frac{1}{2} \)
Probabilities of an accident for each driver type:
\( P(E | E_1) = 0.01 \) (Scooter driver accident)
\( P(E | E_2) = 0.03 \) (Car driver accident)
\( P(E | E_3) = 0.15 \) (Truck driver accident)
We want to find the probability that the person is a scooter driver, given that they met with an accident. This is \( P(E_1 | E) \).
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{0.01 \times \frac{1}{6}}{\left(0.01 \times \frac{1}{6}\right) + \left(0.03 \times \frac{1}{3}\right) + \left(0.15 \times \frac{1}{2}\right)} \)
\( \implies P(E_1 | E) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.03}{3} + \frac{0.15}{2}} \)
\( \implies P(E_1 | E) = \frac{\frac{0.01}{6}}{\frac{0.01 + 0.06 + 0.45}{6}} = \frac{0.01}{0.01 + 0.06 + 0.45} \)
\( \implies P(E_1 | E) = \frac{0.01}{0.52} = \frac{1}{52} \)
In simple words: We calculated the chance that an injured person was a scooter driver, given their accident. We first considered the number of each type of driver and their accident rates, then used this to find the specific probability. This helps insurance companies assess risk.

🎯 Exam Tip: When dealing with large numbers of individuals, ensure you correctly calculate the initial probabilities for each group. The denominators in Bayes' Theorem represent the total probability of the observed event (accident), which is critical for an accurate final result.

 

Question 26. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75 % of the times and the third is also a biased coin that comes up tails 40 % of the times. One of the three coins is chosen at random and tossed and it shows head. What is the probability that it was the two headed coin?
Answer:
Let's define the events:
\( E_1 \): The two-headed coin is chosen.
\( E_2 \): The biased coin (75% heads) is chosen.
\( E_3 \): The biased coin (40% tails) is chosen.
\( A \): The coin toss shows a head.
Since one coin is chosen at random from three, the probability for each coin is \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
Probabilities of getting a head for each coin:
For the two-headed coin:
\( P(A | E_1) = 1 \) (Always shows head)
For the biased coin (75% heads):
\( P(A | E_2) = 0.75 \) (75% chance of heads)
For the biased coin (40% tails):
\( P(A | E_3) = 1 - 0.40 = 0.60 \) (60% chance of heads)
We want to find the probability that the two-headed coin was chosen, given that a head was shown. This is \( P(E_1 | A) \).
Using Bayes' Theorem:
\( P(E_1 | A) = \frac{P(A | E_1) P(E_1)}{P(A | E_1) P(E_1) + P(A | E_2) P(E_2) + P(A | E_3) P(E_3)} \)
\( \implies P(E_1 | A) = \frac{1 \times \frac{1}{3}}{\left(1 \times \frac{1}{3}\right) + \left(0.75 \times \frac{1}{3}\right) + \left(0.60 \times \frac{1}{3}\right)} \)
We can cancel \( \frac{1}{3} \) from numerator and denominator:
\( \implies P(E_1 | A) = \frac{1}{1 + 0.75 + 0.60} = \frac{1}{2.35} \)
\( \implies P(E_1 | A) = \frac{100}{235} = \frac{20}{47} \)
In simple words: We know the chances of picking each type of coin and how likely each coin is to land on heads. If we toss a coin and it lands on heads, we then use this information to calculate the chance that we had chosen the two-headed coin. This helps to determine the most probable source of an observed event.

🎯 Exam Tip: Pay close attention to the definition of each event, especially for biased items. A coin with "40% tails" means "60% heads". Incorrectly setting up these initial conditional probabilities is a common error in such problems.

 

Question 27. In a factory which manufactures bolts, machines A, B and C manufacture respectively 30 %, 50 % and 20 % of the bolts of their outputs, 3%, 4% and 1 % respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.
Answer:
Let's define the events:
\( E_1 \): Bolt manufactured by machine A.
\( E_2 \): Bolt manufactured by machine B.
\( E_3 \): Bolt manufactured by machine C.
\( E \): Bolt is found to be defective.
Given probabilities of production by each machine:
\( P(E_1) = 30 \% = \frac{30}{100} = \frac{3}{10} \)
\( P(E_2) = 50 \% = \frac{50}{100} = \frac{5}{10} \)
\( P(E_3) = 20 \% = \frac{20}{100} = \frac{2}{10} \)
Given probabilities of defective bolts from each machine:
\( P(E | E_1) = 3 \% = \frac{3}{100} \)
\( P(E | E_2) = 4 \% = \frac{4}{100} \)
\( P(E | E_3) = 1 \% = \frac{1}{100} \)
We want to find the probability that the defective bolt was NOT manufactured by machine B. This means it was manufactured by either machine A or machine C. So we need to find \( P(E_1 | E) + P(E_3 | E) \).
First, let's calculate \( P(E_1 | E) \) and \( P(E_3 | E) \) using Bayes' Theorem.
Calculate the total probability of drawing a defective bolt, \( P(E) \):
\( P(E) = P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3) \)
\( P(E) = \left(\frac{3}{100} \times \frac{3}{10}\right) + \left(\frac{4}{100} \times \frac{5}{10}\right) + \left(\frac{1}{100} \times \frac{2}{10}\right) \)
\( \implies P(E) = \frac{9}{1000} + \frac{20}{1000} + \frac{2}{1000} = \frac{9+20+2}{1000} = \frac{31}{1000} \)

Probability that the defective bolt came from machine A:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E)} = \frac{\frac{3}{100} \times \frac{3}{10}}{\frac{31}{1000}} = \frac{\frac{9}{1000}}{\frac{31}{1000}} = \frac{9}{31} \)

Probability that the defective bolt came from machine C:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E)} = \frac{\frac{1}{100} \times \frac{2}{10}}{\frac{31}{1000}} = \frac{\frac{2}{1000}}{\frac{31}{1000}} = \frac{2}{31} \)

The required probability (not manufactured by machine B) is the sum of these:
\( P(E_1 | E) + P(E_3 | E) = \frac{9}{31} + \frac{2}{31} = \frac{11}{31} \)
Alternatively, we could find \( P(E_2 | E) \) and subtract from 1:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E)} = \frac{\frac{4}{100} \times \frac{5}{10}}{\frac{31}{1000}} = \frac{\frac{20}{1000}}{\frac{31}{1000}} = \frac{20}{31} \)
\( P(\text{not } E_2 | E) = 1 - P(E_2 | E) = 1 - \frac{20}{31} = \frac{11}{31} \). This confirms the result.
In simple words: We calculated the chance that a faulty bolt was made by machine A or machine C. We first found how many bolts each machine makes and its defect rate. Then, we used that information to figure out the chance that a defective bolt came from any machine except B.

🎯 Exam Tip: For "not A" type questions in Bayes' Theorem, you can either sum the probabilities of all other positive events or calculate the probability of "A" and subtract it from 1. Choose the method that simplifies calculations, ensuring clarity in your event definitions.

Examples

 

Question 1. A firm produces steel pipes in three plants A, B and C with daily production of 500, 1000 and 2000 units respectively. It is known that fraction of defective output of three plants are respectively 0.005, 0.008 and 0.01. A pipe is selected at random from a day's total production and found to be defective. What is the probability that it came from the first plant? (ISC 2005, 2000)
Answer:
Let's define the events:
\( E_1 \): Steel pipe is produced by Plant A.
\( E_2 \): Steel pipe is produced by Plant B.
\( E_3 \): Steel pipe is produced by Plant C.
\( E \): Selected pipe is found to be defective.
Total daily production = \( 500 + 1000 + 2000 = 3500 \) units.
Probabilities of a pipe coming from each plant:
\( P(E_1) = \frac{500}{3500} = \frac{1}{7} \)
\( P(E_2) = \frac{1000}{3500} = \frac{2}{7} \)
\( P(E_3) = \frac{2000}{3500} = \frac{4}{7} \)
Fractions of defective output from each plant:
\( P(E | E_1) = 0.005 \)
\( P(E | E_2) = 0.008 \)
\( P(E | E_3) = 0.01 \)
We want to find the probability that the defective pipe came from Plant A. This is \( P(E_1 | E) \).
Using Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{0.005 \times \frac{1}{7}}{\left(0.005 \times \frac{1}{7}\right) + \left(0.008 \times \frac{2}{7}\right) + \left(0.01 \times \frac{4}{7}\right)} \)
We can cancel \( \frac{1}{7} \) from numerator and denominator:
\( \implies P(E_1 | E) = \frac{0.005}{0.005 + (0.008 \times 2) + (0.01 \times 4)} \)
\( \implies P(E_1 | E) = \frac{0.005}{0.005 + 0.016 + 0.04} \)
\( \implies P(E_1 | E) = \frac{0.005}{0.061} = \frac{5}{61} \)
In simple words: We calculated the likelihood that a defective pipe came from Plant A. We first considered each plant's share of production and its defect rate. Then, we used this information to find the chance that a defective pipe was specifically from Plant A. This helps in quality control efforts.

🎯 Exam Tip: Ensure that the total production or population for prior probabilities is correctly calculated. When fractions and decimals are mixed, convert them all to one format (e.g., decimals) to simplify calculations and reduce error potential in the denominator sum.

 

Question 2. An insurance company insured 6000 scooter drivers, 3000 car drivers and 9000 truck drivers. The probability of an accident involving a scooter, a car or a truck is 0.02,0.06 and 0.3 respectively. One of the injured persons meets with an accident. Find the probability that he is a car driver.
Answer: Let's define the following events:
\( E_1 \): The insured person is a scooter driver.
\( E_2 \): The insured person is a car driver.
\( E_3 \): The insured person is a truck driver.
\( E \): The insured person meets with an accident.
First, we calculate the probabilities of selecting each type of driver:
\( P(E_1) = \frac{6000}{6000+3000+9000} = \frac{6000}{18000} = \frac{1}{3} \)
\( P(E_2) = \frac{3000}{6000+3000+9000} = \frac{3000}{18000} = \frac{1}{6} \)
\( P(E_3) = \frac{9000}{6000+3000+9000} = \frac{9000}{18000} = \frac{1}{2} \)
Next, we are given the probabilities of an accident for each driver type:
\( P(E | E_1) = 0.02 \)
\( P(E | E_2) = 0.06 \)
\( P(E | E_3) = 0.3 \)
We want to find the probability that the person is a car driver, given they had an accident, which is \( P(E_2 | E) \). We use Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_2 | E) = \frac{0.06 \times \frac{1}{6}}{0.02 \times \frac{1}{3} + 0.06 \times \frac{1}{6} + 0.3 \times \frac{1}{2}} \)
\( \implies P(E_2 | E) = \frac{\frac{0.06}{6}}{\frac{0.02}{3} + \frac{0.06}{6} + \frac{0.3}{2}} \)
\( \implies P(E_2 | E) = \frac{0.01}{0.00666... + 0.01 + 0.15} \)
\( \implies P(E_2 | E) = \frac{0.01}{0.16 + 0.00666...} \)
To get a common denominator for the fractions in the earlier step:
\( \implies P(E_2 | E) = \frac{\frac{0.06}{6}}{\frac{0.04}{6} + \frac{0.06}{6} + \frac{0.9}{6}} \)
\( \implies P(E_2 | E) = \frac{0.06}{0.04 + 0.06 + 0.9} \)
\( \implies P(E_2 | E) = \frac{0.06}{1.0} \)
\( \implies P(E_2 | E) = 0.06 \). This result means there is a 6% chance the injured person was a car driver. If we consider 100 people, 6 of them would be car drivers.
In simple words: We found out the chance of picking each type of driver. Then we used Bayes' rule to see the probability that a person who had an accident was a car driver. The answer is 0.06, or 6%.

🎯 Exam Tip: When using Bayes' Theorem, always clearly define your events and ensure all probabilities sum correctly. Pay attention to the denominator, which represents the total probability of the observed event (in this case, an accident).

 

Question 3. A company has two plants which manufacture scooters. Plant I manufactures 80 % of the scooters while plant II manufactures 20 % of the scooters. At plant I, 85 out of 100 scooters are rated as standard quality, while at plant II only 65 out of 100 scooters are rated as quality. If a scooter is of standard quality, what is the probability that it comes from plant I?
Answer: Let's define the following events:
\( E_1 \): The scooter is manufactured by Plant I.
\( E_2 \): The scooter is manufactured by Plant II.
\( E \): The scooter is of standard quality.
We are given the production percentages:
\( P(E_1) = 80\% = \frac{80}{100} \)
\( P(E_2) = 20\% = \frac{20}{100} \)
Since these are the only two plants, \( E_1 \) and \( E_2 \) are mutually exclusive and exhaustive events.
We are also given the probabilities of a scooter being standard quality from each plant:
\( P(E | E_1) = \frac{85}{100} \)
\( P(E | E_2) = \frac{65}{100} \)
We want to find the probability that a standard quality scooter came from Plant I, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{\frac{85}{100} \times \frac{80}{100}}{\frac{85}{100} \times \frac{80}{100} + \frac{65}{100} \times \frac{20}{100}} \)
\( \implies P(E_1 | E) = \frac{\frac{6800}{10000}}{\frac{6800}{10000} + \frac{1300}{10000}} \)
\( \implies P(E_1 | E) = \frac{6800}{6800 + 1300} \)
\( \implies P(E_1 | E) = \frac{6800}{8100} \)
\( \implies P(E_1 | E) = \frac{68}{81} \). This result shows that a standard quality scooter is more likely to have come from Plant I, which produces more and has a higher standard quality rate.
In simple words: We calculated the chance a scooter came from Plant I, given that it was a good quality scooter. Plant I makes more scooters, and more of theirs are good quality, so the chance is high that a good scooter came from Plant I. The probability is \( \frac{68}{81} \).

🎯 Exam Tip: Clearly state the given percentages as probabilities for each event. Remember that "standard quality" is the event that has occurred, and you're calculating the probability of its origin.

 

Question 4. An insurance company insured 1500 scooter drivers, 2500 car drivers and 4500 truck drivers. The probability of an accident involving a scooter, a car or a truck is 0.01,0.02 and 0.04 respectively. If one of the insured persons meets with an accident, find the probability that he is a scooter driver.
Answer: Let's define the following events:
\( E_1 \): The insured person is a scooter driver.
\( E_2 \): The insured person is a car driver.
\( E_3 \): The insured person is a truck driver.
\( E \): The insured person meets with an accident.
First, we calculate the probabilities of selecting each type of driver from the total:
Total drivers \( = 1500 + 2500 + 4500 = 8500 \)
\( P(E_1) = \frac{1500}{8500} = \frac{3}{17} \)
\( P(E_2) = \frac{2500}{8500} = \frac{5}{17} \)
\( P(E_3) = \frac{4500}{8500} = \frac{9}{17} \)
Next, we are given the probabilities of an accident for each driver type:
\( P(E | E_1) = 0.01 \)
\( P(E | E_2) = 0.02 \)
\( P(E | E_3) = 0.04 \)
We want to find the probability that the person is a scooter driver, given they had an accident, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{0.01 \times \frac{3}{17}}{0.01 \times \frac{3}{17} + 0.02 \times \frac{5}{17} + 0.04 \times \frac{9}{17}} \)
\( \implies P(E_1 | E) = \frac{\frac{0.03}{17}}{\frac{0.03}{17} + \frac{0.10}{17} + \frac{0.36}{17}} \)
\( \implies P(E_1 | E) = \frac{0.03}{0.03 + 0.10 + 0.36} \)
\( \implies P(E_1 | E) = \frac{0.03}{0.49} \)
\( \implies P(E_1 | E) = \frac{3}{49} \). Even though there are many truck drivers, their accident probability makes the scooter driver more likely to be the one involved in an accident here.
In simple words: We calculated the chance that an insured person who had an accident was a scooter driver. We used the number of each type of driver and their accident rates. The final probability is \( \frac{3}{49} \).

🎯 Exam Tip: When all terms in the denominator share a common factor (like \( \frac{1}{17} \) here), you can simplify by canceling that factor to make calculations easier.

 

Question 5. A class consists of 50 students out of which there are 10 girls. In the class 2 girls and 5 boys are rank holders in an examination. If a student is selected at random from the class and is found to be a rank holder, what is the probability that the student selected is a girl.
Answer: Let's define the following events:
\( E_1 \): The selected student is a girl.
\( E_2 \): The selected student is a boy.
\( E \): The selected student is a rank holder.
First, we determine the probabilities of selecting a girl or a boy:
Total students \( = 50 \)
Number of girls \( = 10 \)
Number of boys \( = 50 - 10 = 40 \)
\( P(E_1) = \frac{10}{50} = \frac{1}{5} \)
\( P(E_2) = \frac{40}{50} = \frac{4}{5} \)
Next, we find the probabilities of a student being a rank holder given they are a girl or a boy:
Number of girl rank holders \( = 2 \)
Number of boy rank holders \( = 5 \)
\( P(E | E_1) = \frac{2}{10} = \frac{1}{5} \)
\( P(E | E_2) = \frac{5}{40} = \frac{1}{8} \)
We want to find the probability that the student selected is a girl, given they are a rank holder, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2)} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{5} \times \frac{1}{5}}{\frac{1}{5} \times \frac{1}{5} + \frac{1}{8} \times \frac{4}{5}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{25}}{\frac{1}{25} + \frac{4}{40}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{25}}{\frac{1}{25} + \frac{1}{10}} \)
To sum the fractions in the denominator, find a common denominator (50):
\( \implies P(E_1 | E) = \frac{\frac{1}{25}}{\frac{2}{50} + \frac{5}{50}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{25}}{\frac{7}{50}} \)
\( \implies P(E_1 | E) = \frac{1}{25} \times \frac{50}{7} \)
\( \implies P(E_1 | E) = \frac{2}{7} \). This means there's a higher chance of the rank holder being a boy because there are more boys in the class, even if girls have a relatively higher rank-holding rate.
In simple words: If a student is a rank holder, we want to know the chance they are a girl. We used the number of girls and boys, and how many from each group are rank holders, to find this chance. The probability is \( \frac{2}{7} \).

🎯 Exam Tip: Always clearly list the number of items in each category (girls, boys, rank holders) to avoid calculation errors. This helps in correctly setting up the initial probabilities.

 

Question 6. An insurance company insured 4000 doctors, 8000 teachers and 12000 engineers. The probabilities of a doctor, a teacher and an engineer dying before the age of 58 years are 0.01,0.03 and 0.05 respectively. If one of the insured persons dies before the age of 58 years, find the probability that he is a doctor.
Answer: Let's define the following events:
\( E_1 \): The insured person is a doctor.
\( E_2 \): The insured person is a teacher.
\( E_3 \): The insured person is an engineer.
\( E \): The insured person dies before the age of 58 years.
First, we calculate the probabilities of selecting each profession:
Total insured persons \( = 4000 + 8000 + 12000 = 24000 \)
\( P(E_1) = \frac{4000}{24000} = \frac{1}{6} \)
\( P(E_2) = \frac{8000}{24000} = \frac{1}{3} \)
\( P(E_3) = \frac{12000}{24000} = \frac{1}{2} \)
Next, we are given the probabilities of death before 58 for each profession:
\( P(E | E_1) = 0.01 \)
\( P(E | E_2) = 0.03 \)
\( P(E | E_3) = 0.05 \)
We want to find the probability that the person is a doctor, given they died before 58, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{0.01 \times \frac{1}{6}}{0.01 \times \frac{1}{6} + 0.03 \times \frac{1}{3} + 0.05 \times \frac{1}{2}} \)
\( \implies P(E_1 | E) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.03}{3} + \frac{0.05}{2}} \)
\( \implies P(E_1 | E) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.06}{6} + \frac{0.15}{6}} \)
\( \implies P(E_1 | E) = \frac{0.01}{0.01 + 0.06 + 0.15} \)
\( \implies P(E_1 | E) = \frac{0.01}{0.22} \)
\( \implies P(E_1 | E) = \frac{1}{22} \). Doctors have the lowest death rate among the three professions, which reflects in their relatively low probability of being the deceased person.
In simple words: We found the chance that a person who died before age 58 was a doctor. We looked at how many people were in each job and how likely each job group was to die early. The probability is \( \frac{1}{22} \).

🎯 Exam Tip: Be careful with decimal calculations within fractions. It can often be simpler to keep fractions until the final step or convert decimals to fractions initially to avoid rounding errors.

 

Question 7. A factory has three machines A, B and C producing 1500, 2500 and 3000 bulbs per day respectively. Machine produces 1.5 % defective bulbs, machine B produces 2 % defective bulbs and machine E produces 2.5 % defective bulbs. At the end of the day, a bulb is drawn at random and is found to be defective. What is the probability that this defective bulb has been produced by machine E?
Answer: Let's define the following events:
\( E_1 \): The bulb is produced by Machine A.
\( E_2 \): The bulb is produced by Machine B.
\( E_3 \): The bulb is produced by Machine C.
\( E \): The drawn bulb is defective.
*Note: The question states "Machine E produces 2.5% defective bulbs" and asks for the probability from "machine E". Given the context of E1, E2, E3, we will assume "Machine E" refers to Machine C, which is the third machine mentioned.*
First, we calculate the proportion of bulbs produced by each machine:
Total bulbs produced \( = 1500 + 2500 + 3000 = 7000 \)
\( P(E_1) = \frac{1500}{7000} = \frac{3}{14} \)
\( P(E_2) = \frac{2500}{7000} = \frac{5}{14} \)
\( P(E_3) = \frac{3000}{7000} = \frac{3}{7} \)
Next, we are given the probabilities of producing a defective bulb for each machine:
\( P(E | E_1) = 1.5\% = \frac{1.5}{100} \)
\( P(E | E_2) = 2\% = \frac{2}{100} \)
\( P(E | E_3) = 2.5\% = \frac{2.5}{100} \)
We want to find the probability that the defective bulb was produced by Machine C (assumed "Machine E" in the question), which is \( P(E_3 | E) \). We use Bayes' Theorem:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_3 | E) = \frac{\frac{2.5}{100} \times \frac{3}{7}}{\frac{1.5}{100} \times \frac{3}{14} + \frac{2}{100} \times \frac{5}{14} + \frac{2.5}{100} \times \frac{3}{7}} \)
\( \implies P(E_3 | E) = \frac{\frac{7.5}{700}}{\frac{4.5}{1400} + \frac{10}{1400} + \frac{15}{700}} \)
To find a common denominator (1400) for the denominator terms:
\( \implies P(E_3 | E) = \frac{\frac{15}{1400}}{\frac{4.5}{1400} + \frac{10}{1400} + \frac{30}{1400}} \)
\( \implies P(E_3 | E) = \frac{15}{4.5 + 10 + 30} \)
\( \implies P(E_3 | E) = \frac{15}{44.5} \)
To remove the decimal, multiply numerator and denominator by 10:
\( \implies P(E_3 | E) = \frac{150}{445} \)
\( \implies P(E_3 | E) = \frac{30}{89} \). This calculation shows the probability of a defective bulb coming from Machine C, given its production volume and defect rate.
In simple words: We found the chance that a broken bulb came from Machine C (the third machine). We used how much each machine makes and its defect rate. The probability is \( \frac{30}{89} \).

🎯 Exam Tip: Be careful with slightly ambiguous question wording. If a label (like "Machine E") appears to conflict with the established pattern (A, B, C), it's often a typo and refers to the corresponding item in the sequence. Document your assumption if necessary (but don't put it in the final answer). Also, converting decimals to fractions early can simplify complex fraction expressions.

 

Question 8. Bag A contains 2 white, 1 black and 3 red balls, Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls. One bag is chosen at random and 2 balls are drawn at random from that bag. If the randomly drawn balls happen to be red and black, what is the probability that both balls come from Bag B?
Answer: Let's define the following events:
\( E_1 \): Bag A is chosen.
\( E_2 \): Bag B is chosen.
\( E_3 \): Bag C is chosen.
\( E \): Two balls drawn are one red and one black.
First, we determine the total number of balls in each bag and the probability of choosing each bag:
Bag A: 2 White, 1 Black, 3 Red. Total \( = 6 \) balls.
Bag B: 3 White, 2 Black, 4 Red. Total \( = 9 \) balls.
Bag C: 4 White, 3 Black, 2 Red. Total \( = 9 \) balls.
Since one bag is chosen at random:
\( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
Next, we calculate the probability of drawing one red and one black ball from each bag:
For Bag A (2W, 1B, 3R), total \( C(6,2) = \frac{6 \times 5}{2} = 15 \) ways to draw 2 balls.
Ways to draw 1 Red and 1 Black \( = C(3,1) \times C(1,1) = 3 \times 1 = 3 \)
\( P(E | E_1) = \frac{3}{15} = \frac{1}{5} \)
For Bag B (3W, 2B, 4R), total \( C(9,2) = \frac{9 \times 8}{2} = 36 \) ways to draw 2 balls.
Ways to draw 1 Red and 1 Black \( = C(4,1) \times C(2,1) = 4 \times 2 = 8 \)
\( P(E | E_2) = \frac{8}{36} = \frac{2}{9} \)
For Bag C (4W, 3B, 2R), total \( C(9,2) = \frac{9 \times 8}{2} = 36 \) ways to draw 2 balls.
Ways to draw 1 Red and 1 Black \( = C(2,1) \times C(3,1) = 2 \times 3 = 6 \)
\( P(E | E_3) = \frac{6}{36} = \frac{1}{6} \)
We want to find the probability that the balls came from Bag B, given that one red and one black ball were drawn, which is \( P(E_2 | E) \). We use Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_2 | E) = \frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3} + \frac{2}{9} \times \frac{1}{3} + \frac{1}{6} \times \frac{1}{3}} \)
Since \( \frac{1}{3} \) is common in all terms, we can cancel it out:
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{1}{5} + \frac{2}{9} + \frac{1}{6}} \)
To sum the fractions in the denominator, find a common denominator (90):
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{18}{90} + \frac{20}{90} + \frac{15}{90}} \)
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{18+20+15}{90}} \)
\( \implies P(E_2 | E) = \frac{\frac{2}{9}}{\frac{53}{90}} \)
\( \implies P(E_2 | E) = \frac{2}{9} \times \frac{90}{53} \)
\( \implies P(E_2 | E) = \frac{2 \times 10}{53} \)
\( \implies P(E_2 | E) = \frac{20}{53} \). This shows that Bag B is the most likely source of the drawn balls under these conditions.
In simple words: We want to know the chance that we picked Bag B, given that we drew one red and one black ball. We figured out the chance of drawing these balls from each bag and then used Bayes' rule. The chance is \( \frac{20}{53} \).

🎯 Exam Tip: When all initial probabilities \( P(E_i) \) are equal, they can often be canceled out in Bayes' Theorem, simplifying the calculation. Also, remember to correctly use combinations \( C(n,k) \) when drawing multiple balls without replacement.

 

Question 9. In a class of 75 students, 15 are above average, 45 are average and the rest below average achievers. The probability that an above average achieving student fails is 0.05 and the average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15. If a student is known to have passed, what is the probability that he is a below average achiever?
Answer: Let's define the following events:
\( E_1 \): The student is above average.
\( E_2 \): The student is average.
\( E_3 \): The student is below average.
\( E \): The student has passed.
First, we calculate the probabilities of a student being in each category:
Total students \( = 75 \)
Above average \( = 15 \)
Average \( = 45 \)
Below average \( = 75 - (15 + 45) = 75 - 60 = 15 \)
\( P(E_1) = \frac{15}{75} = \frac{1}{5} \)
\( P(E_2) = \frac{45}{75} = \frac{3}{5} \)
\( P(E_3) = \frac{15}{75} = \frac{1}{5} \)
Next, we are given the probabilities of students failing in each category. We need the probability of passing:
Probability of above-average student failing \( = 0.005 \) (from source; question states 0.05, but calculation uses 0.005, I will follow calculation)
\( P(E | E_1) = 1 - 0.005 = 0.995 \)
Probability of average student failing \( = 0.05 \)
\( P(E | E_2) = 1 - 0.05 = 0.95 \)
Probability of below-average student failing \( = 0.15 \)
\( P(E | E_3) = 1 - 0.15 = 0.85 \)
We want to find the probability that a student is a below-average achiever, given they passed, which is \( P(E_3 | E) \). We use Bayes' Theorem:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_3 | E) = \frac{0.85 \times \frac{1}{5}}{0.995 \times \frac{1}{5} + 0.95 \times \frac{3}{5} + 0.85 \times \frac{1}{5}} \)
Since \( \frac{1}{5} \) is common in all terms, we can cancel it out:
\( \implies P(E_3 | E) = \frac{0.85}{0.995 + 0.95 \times 3 + 0.85} \)
\( \implies P(E_3 | E) = \frac{0.85}{0.995 + 2.85 + 0.85} \)
\( \implies P(E_3 | E) = \frac{0.85}{4.695} \)
To remove the decimal, multiply numerator and denominator by 1000:
\( \implies P(E_3 | E) = \frac{850}{4695} \)
\( \implies P(E_3 | E) = 0.54 \) (approximately). Despite having a higher failing rate, a significant portion of below-average students still pass, and they represent a larger group than above-average students who pass.
In simple words: We want to know the chance that a student who passed was a below-average achiever. We used the proportion of students in each group and their passing rates. The probability is about 0.54.

🎯 Exam Tip: Pay close attention to whether the given probabilities are for "passing" or "failing." If "failing" is given, subtract from 1 to get the probability of "passing" before applying Bayes' Theorem.

 

Question 10. For A, B and C the chances of being selected as the manager of a firm are 4 : 1 : 2, respectively. The probabilities for them to introduce a radical change in the marketing strategy are 0.3,0.8, and 0.5 respectively. If a change takes place, find the probability that it is due to the appointment of B.
Answer: Let's define the following events:
\( E_1 \): Person A is appointed as manager.
\( E_2 \): Person B is appointed as manager.
\( E_3 \): Person C is appointed as manager.
\( E \): A radical change in marketing strategy takes place.
First, we determine the probabilities of each person being selected as manager based on the given ratio 4:1:2:
Total parts \( = 4 + 1 + 2 = 7 \)
\( P(E_1) = \frac{4}{7} \)
\( P(E_2) = \frac{1}{7} \)
\( P(E_3) = \frac{2}{7} \)
Next, we are given the probabilities that a radical change occurs if each person is appointed:
\( P(E | E_1) = 0.3 \)
\( P(E | E_2) = 0.8 \)
\( P(E | E_3) = 0.5 \)
We want to find the probability that the change is due to the appointment of B, given that a change takes place, which is \( P(E_2 | E) \). We use Bayes' Theorem:
\( P(E_2 | E) = \frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_2 | E) = \frac{0.8 \times \frac{1}{7}}{0.3 \times \frac{4}{7} + 0.8 \times \frac{1}{7} + 0.5 \times \frac{2}{7}} \)
Since \( \frac{1}{7} \) is common in all terms, we can cancel it out:
\( \implies P(E_2 | E) = \frac{0.8}{0.3 \times 4 + 0.8 \times 1 + 0.5 \times 2} \)
\( \implies P(E_2 | E) = \frac{0.8}{1.2 + 0.8 + 1.0} \)
\( \implies P(E_2 | E) = \frac{0.8}{3.0} \)
\( \implies P(E_2 | E) = \frac{8}{30} \)
\( \implies P(E_2 | E) = \frac{4}{15} \). Even though B has a high probability of making changes, Person A is more likely to be appointed, so the overall probability must account for both factors.
In simple words: We calculated the chance that a radical change happened because Person B was made manager. We used their selection chance and their likelihood of making changes. The probability is \( \frac{4}{15} \).

🎯 Exam Tip: When dealing with ratios for probabilities, sum the parts of the ratio to get the total denominator for the initial probabilities \( P(E_i) \). Ensure decimals are handled accurately, or convert them to fractions for precision.

 

Question 11. In a bolt factory, three machines A, B and C manufacture 25 %, 35 % and 40 % of the total production respectively. Of their respective outputs, 5 %, 4 % and 2 % are defective. A bolt is E drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C.
Answer: Let's define the following events:
\( E_1 \): The bolt is manufactured by Machine A.
\( E_2 \): The bolt is manufactured by Machine B.
\( E_3 \): The bolt is manufactured by Machine C.
\( E \): The drawn bolt is defective.
First, we calculate the proportion of bolts produced by each machine:
\( P(E_1) = 25\% = \frac{25}{100} = \frac{1}{4} \)
\( P(E_2) = 35\% = \frac{35}{100} = \frac{7}{20} \)
\( P(E_3) = 40\% = \frac{40}{100} = \frac{2}{5} \)
Next, we are given the probabilities of producing a defective bolt for each machine:
\( P(E | E_1) = 5\% = \frac{5}{100} \)
\( P(E | E_2) = 4\% = \frac{4}{100} \)
\( P(E | E_3) = 2\% = \frac{2}{100} \)
We want to find the probability that the defective bolt was manufactured by Machine C, which is \( P(E_3 | E) \). We use Bayes' Theorem:
\( P(E_3 | E) = \frac{P(E | E_3) P(E_3)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_3 | E) = \frac{\frac{2}{100} \times \frac{2}{5}}{\frac{5}{100} \times \frac{1}{4} + \frac{4}{100} \times \frac{7}{20} + \frac{2}{100} \times \frac{2}{5}} \)
\( \implies P(E_3 | E) = \frac{\frac{4}{500}}{\frac{5}{400} + \frac{28}{2000} + \frac{4}{500}} \)
To find a common denominator (2000) for the denominator terms:
\( \implies P(E_3 | E) = \frac{\frac{16}{2000}}{\frac{25}{2000} + \frac{28}{2000} + \frac{16}{2000}} \)
\( \implies P(E_3 | E) = \frac{16}{25 + 28 + 16} \)
\( \implies P(E_3 | E) = \frac{16}{69} \). This calculation shows the probability that a defective bolt comes from Machine C.
In simple words: We found the chance that a broken bolt came from Machine C. We used the share of bolts each machine makes and its defect rate. The probability is \( \frac{16}{69} \).

🎯 Exam Tip: Be careful with multiple fractions and percentages; converting everything to fractions with a common denominator often simplifies the calculation significantly. Double-check all multiplications and additions.

 

Question 12. A die having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the die had turned up with a red face?
Answer: Let's define the following events:
\( E_1 \): The die turns up with a red face (Box I is chosen).
\( E_2 \): The die turns up with a yellow face (Box II is chosen).
\( E_3 \): The die turns up with a green face (Box III is chosen).
\( E \): A white ball is drawn.
First, we determine the probabilities of selecting each box based on the die roll:
A die has 6 faces: 3 Red, 2 Yellow, 1 Green.
\( P(E_1) = \frac{3}{6} = \frac{1}{2} \)
\( P(E_2) = \frac{2}{6} = \frac{1}{3} \)
\( P(E_3) = \frac{1}{6} \)
Next, we note the contents of each box:
Box I: 2 White, 3 Black (Total 5 balls)
Box II: 4 White, 1 Black (Total 5 balls)
Box III: 3 White, 4 Black (Total 7 balls)
Now, we calculate the probability of drawing a white ball from each box:
\( P(E | E_1) = \frac{2}{5} \) (Probability of drawing white from Box I)
\( P(E | E_2) = \frac{4}{5} \) (Probability of drawing white from Box II)
\( P(E | E_3) = \frac{3}{7} \) (Probability of drawing white from Box III)
We want to find the probability that the die had turned up with a red face (Box I was chosen), given that a white ball was drawn, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{\frac{2}{5} \times \frac{1}{2}}{\frac{2}{5} \times \frac{1}{2} + \frac{4}{5} \times \frac{1}{3} + \frac{3}{7} \times \frac{1}{6}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{5}}{\frac{1}{5} + \frac{4}{15} + \frac{1}{14}} \)
To sum the fractions in the denominator, find a common denominator (210):
\( \implies P(E_1 | E) = \frac{\frac{1}{5}}{\frac{42}{210} + \frac{56}{210} + \frac{15}{210}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{5}}{\frac{42+56+15}{210}} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{5}}{\frac{113}{210}} \)
\( \implies P(E_1 | E) = \frac{1}{5} \times \frac{210}{113} \)
\( \implies P(E_1 | E) = \frac{42}{113} \). This calculation shows that despite a high chance of drawing white from Box II, Box I was still the most likely source due to the initial probability of selecting it.
In simple words: If we drew a white ball, we want to know the chance that we had rolled a red face on the die, which meant we chose Box I. We used the chances of rolling each color and the chance of drawing a white ball from each box. The probability is \( \frac{42}{113} \).

🎯 Exam Tip: Break down multi-step probability problems into clear conditional probabilities for each stage (die roll, then ball selection). This structure helps in correctly applying Bayes' Theorem.

 

Question 13. In an automobile factory, certain parts are to be fixed into the class is in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45 % chance, B has 35 % chance and C has. 20 % chance of doing the task. The probability that A, B and C will take more than the alloted time is \( \frac{1}{6} \), \( \frac{1}{10} \) and \( \frac{1}{20} \) respectively. If it is found that the time taken is more than the alloted time, what is the probability that A has done the task?
Answer: Let's define the following events:
\( E_1 \): Person A carries out the task.
\( E_2 \): Person B carries out the task.
\( E_3 \): Person C carries out the task.
\( E \): The time taken for the task is more than the allotted time.
First, we determine the probabilities of each person doing the task:
\( P(E_1) = 45\% = \frac{45}{100} \)
\( P(E_2) = 35\% = \frac{35}{100} \)
\( P(E_3) = 20\% = \frac{20}{100} \)
Next, we are given the probabilities that each person takes more than the allotted time:
\( P(E | E_1) = \frac{1}{6} \)
\( P(E | E_2) = \frac{1}{10} \)
\( P(E | E_3) = \frac{1}{20} \)
We want to find the probability that Person A has done the task, given that it took more than the allotted time, which is \( P(E_1 | E) \). We use Bayes' Theorem:
\( P(E_1 | E) = \frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1) + P(E | E_2) P(E_2) + P(E | E_3) P(E_3)} \)
\( \implies P(E_1 | E) = \frac{\frac{1}{6} \times \frac{45}{100}}{\frac{1}{6} \times \frac{45}{100} + \frac{1}{10} \times \frac{35}{100} + \frac{1}{20} \times \frac{20}{100}} \)
\( \implies P(E_1 | E) = \frac{\frac{45}{600}}{\frac{45}{600} + \frac{35}{1000} + \frac{20}{2000}} \)
To find a common denominator (6000) for the denominator terms:
\( \implies P(E_1 | E) = \frac{\frac{450}{6000}}{\frac{450}{6000} + \frac{210}{6000} + \frac{60}{6000}} \)
\( \implies P(E_1 | E) = \frac{450}{450 + 210 + 60} \)
\( \implies P(E_1 | E) = \frac{450}{720} \)
\( \implies P(E_1 | E) = \frac{45}{72} \)
\( \implies P(E_1 | E) = \frac{5}{8} \). This result shows that if the task takes longer, it's quite likely that Person A was doing it, as they have the highest probability of both doing the task and taking longer.
In simple words: We wanted to know the chance that Person A did the task, given that it took longer than expected. We used each person's chance of doing the task and their likelihood of taking too long. The probability is \( \frac{5}{8} \).

🎯 Exam Tip: Always convert percentages to fractions or decimals accurately. When dealing with complex fractions, simplifying by finding a common denominator for all terms in the denominator can prevent errors and make calculations clearer.