RBSE Solutions Class 12 Maths Chapter 16 Probability and Probability Distribution Exercise 16.3

Question 1. Given there are two bags I and II. Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II.
Answer: Let \( E_1 \) be the event that the ball is drawn from Bag I. Let \( E_2 \) be the event that the ball is drawn from Bag II. Let \( A \) be the event that the drawn ball is red.
Since the bags are equally likely to be chosen,
\( P(E_1) = P(E_2) = \frac{1}{2} \)

Bag I has 3 red and 4 black balls, so a total of \( 3+4=7 \) balls.
Probability of drawing a red ball from Bag I is \( P\left(\frac{A}{E_1}\right) = \frac{3}{7} \)

Bag II has 5 red and 6 black balls, so a total of \( 5+6=11 \) balls.
Probability of drawing a red ball from Bag II is \( P\left(\frac{A}{E_2}\right) = \frac{5}{11} \)

We need to find the probability that the red ball was drawn from Bag II, which is \( P\left(\frac{E_2}{A}\right) \). We use Bayes' Theorem:
\[ P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Now, substitute the values:
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{1}{2} \cdot \frac{5}{11}}{\frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{11}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} \]
To add the fractions in the denominator, find a common denominator, which is 154:
\( \frac{3}{14} = \frac{3 \cdot 11}{14 \cdot 11} = \frac{33}{154} \)
\( \frac{5}{22} = \frac{5 \cdot 7}{22 \cdot 7} = \frac{35}{154} \)

\( \implies \)
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{5}{22}}{\frac{33}{154} + \frac{35}{154}} = \frac{\frac{5}{22}}{\frac{33+35}{154}} = \frac{\frac{5}{22}}{\frac{68}{154}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{5}{22} \cdot \frac{154}{68} \]
We can simplify 154 by 22: \( 154 \div 22 = 7 \).
\[ P\left(\frac{E_2}{A}\right) = \frac{5 \cdot 7}{68} = \frac{35}{68} \]
So, the probability that the red ball was drawn from Bag II is \( \frac{35}{68} \). Understanding Bayes' theorem helps to calculate conditional probabilities based on prior knowledge.
In simple words: We want to find the chance that a red ball came from Bag 2, knowing it is red. We use a special formula (Bayes' Theorem) that combines the chance of picking Bag 2 and the chance of getting red from Bag 2, divided by the total chance of getting a red ball from either bag. The final chance is \( \frac{35}{68} \).

🎯 Exam Tip: Clearly define your events \( E_1, E_2, A \) and write down their probabilities before applying Bayes' Theorem. This helps organize your thoughts and avoids calculation errors.

Question 2. A doctor has to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively \( \frac{3}{10} \), \( \frac{1}{5} \), \( \frac{1}{10} \) and \( \frac{1}{5} \). The probabilities that he will be late are \( \frac{1}{4} \), \( \frac{1}{3} \) and \( \frac{1}{12} \), if he comes by train, bus and scooter respectively, but if he comes by other means of transport, than he will not be late. When he arrives, he is late. What is the probability that he comes by train ?
Answer: Let \( E \) be the event that the doctor comes late to visit a patient.
Let \( T_1 \) be the event that the doctor comes by train.
Let \( T_2 \) be the event that the doctor comes by bus.
Let \( T_3 \) be the event that the doctor comes by scooter.
Let \( T_4 \) be the event that the doctor comes by other means of transport.

Given probabilities of travel means:
\( P(T_1) = \frac{3}{10} \)
\( P(T_2) = \frac{1}{5} \)
\( P(T_3) = \frac{1}{10} \)
\( P(T_4) = \frac{1}{5} \)

Given probabilities of being late:
If by train: \( P\left(\frac{E}{T_1}\right) = \frac{1}{4} \)
If by bus: \( P\left(\frac{E}{T_2}\right) = \frac{1}{3} \)
If by scooter: \( P\left(\frac{E}{T_3}\right) = \frac{1}{12} \)
If by other means: \( P\left(\frac{E}{T_4}\right) = 0 \) (because he will not be late)

We need to find the probability that the doctor comes by train, given that he is late. This is \( P\left(\frac{T_1}{E}\right) \). We use Bayes' Theorem:
\[ P\left(\frac{T_1}{E}\right) = \frac{P(T_1) \cdot P\left(\frac{E}{T_1}\right)}{P(T_1) \cdot P\left(\frac{E}{T_1}\right) + P(T_2) \cdot P\left(\frac{E}{T_2}\right) + P(T_3) \cdot P\left(\frac{E}{T_3}\right) + P(T_4) \cdot P\left(\frac{E}{T_4}\right)} \]
Now, substitute the values:
\[ P\left(\frac{T_1}{E}\right) = \frac{\frac{3}{10} \cdot \frac{1}{4}}{\frac{3}{10} \cdot \frac{1}{4} + \frac{1}{5} \cdot \frac{1}{3} + \frac{1}{10} \cdot \frac{1}{12} + \frac{1}{5} \cdot 0} \]
\[ P\left(\frac{T_1}{E}\right) = \frac{\frac{3}{40}}{\frac{3}{40} + \frac{1}{15} + \frac{1}{120} + 0} \]
Find a common denominator for the denominator: 120.
\( \frac{3}{40} = \frac{3 \cdot 3}{40 \cdot 3} = \frac{9}{120} \)
\( \frac{1}{15} = \frac{1 \cdot 8}{15 \cdot 8} = \frac{8}{120} \)
\( \frac{1}{120} = \frac{1}{120} \)

\( \implies \)
\[ P\left(\frac{T_1}{E}\right) = \frac{\frac{3}{40}}{\frac{9}{120} + \frac{8}{120} + \frac{1}{120}} = \frac{\frac{3}{40}}{\frac{9+8+1}{120}} = \frac{\frac{3}{40}}{\frac{18}{120}} \]
\[ P\left(\frac{T_1}{E}\right) = \frac{3}{40} \cdot \frac{120}{18} \]
We can simplify 120 by 40: \( 120 \div 40 = 3 \).
\[ P\left(\frac{T_1}{E}\right) = \frac{3 \cdot 3}{18} = \frac{9}{18} = \frac{1}{2} \]
The probability that the doctor came by train, given he was late, is \( \frac{1}{2} \). Bayes' theorem is a powerful tool to update probabilities based on new evidence.
In simple words: We want to know the chance the doctor came by train, given that he was late. We look at how likely he is to use each transport and how likely he is to be late with each. Using these numbers, we find that there's a 1 in 2 chance he came by train.

🎯 Exam Tip: Carefully list all the given probabilities for each event and conditional event. This organization helps prevent errors when substituting values into Bayes' Theorem formula.

Question 3. Bag I contains 3 red and 5 black balls and bag II contains 4 red and 5 black balls. One ball transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer: Let \( E_1 \) be the event that a red ball is transferred from Bag I to Bag II.
Let \( E_2 \) be the event that a black ball is transferred from Bag I to Bag II.
Let \( A \) be the event that a red ball is drawn from Bag II.

Bag I contains 3 red and 5 black balls, totaling \( 3+5=8 \) balls.
Probability of transferring a red ball: \( P(E_1) = \frac{3}{8} \)
Probability of transferring a black ball: \( P(E_2) = \frac{5}{8} \)

Now, consider the composition of Bag II and the probability of drawing a red ball from it, based on what was transferred.

Case 1: A red ball was transferred (event \( E_1 \)).
Bag II initially has 4 red and 5 black balls (total 9). After transferring a red ball, it will have \( 4+1=5 \) red and 5 black balls (total 10).
Probability of drawing a red ball from Bag II in this case: \( P\left(\frac{A}{E_1}\right) = \frac{5}{10} = \frac{1}{2} \)

Case 2: A black ball was transferred (event \( E_2 \)).
Bag II initially has 4 red and 5 black balls (total 9). After transferring a black ball, it will have 4 red and \( 5+1=6 \) black balls (total 10).
Probability of drawing a red ball from Bag II in this case: \( P\left(\frac{A}{E_2}\right) = \frac{4}{10} = \frac{2}{5} \)

We need to find the probability that the transferred ball was black, given that the ball drawn from Bag II is red. This is \( P\left(\frac{E_2}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{5}{8} \cdot \frac{2}{5}}{\frac{3}{8} \cdot \frac{1}{2} + \frac{5}{8} \cdot \frac{2}{5}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{10}{40}}{\frac{3}{16} + \frac{10}{40}} \]
Simplify the fractions: \( \frac{10}{40} = \frac{1}{4} \) and \( \frac{10}{40} = \frac{1}{4} \).
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{1}{4}}{\frac{3}{16} + \frac{1}{4}} \]
For the denominator, find a common denominator (16): \( \frac{1}{4} = \frac{4}{16} \).
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{1}{4}}{\frac{3}{16} + \frac{4}{16}} = \frac{\frac{1}{4}}{\frac{7}{16}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{1}{4} \cdot \frac{16}{7} = \frac{16}{28} = \frac{4}{7} \]
Thus, the probability that the transferred ball was black, given that the ball drawn from Bag II is red, is \( \frac{4}{7} \). This problem showcases how the initial transfer affects the subsequent probability calculations.
In simple words: We want to know the chance that a black ball was moved from Bag 1 to Bag 2, given that we later drew a red ball from Bag 2. We calculate the chances for two scenarios: a red ball was moved, or a black ball was moved. Using these, and the final observation (red ball drawn), we find the chance that a black ball was transferred is \( \frac{4}{7} \).

🎯 Exam Tip: When dealing with ball transfer problems, always adjust the number of balls in the receiving bag accurately for each transfer scenario before calculating subsequent probabilities.

Question 4. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer: Let \( E_1 \) be the event that Bag I is selected.
Let \( E_2 \) be the event that Bag II is selected.
Let \( A \) be the event that a red ball is drawn.

Since one of the two bags is selected at random,
\( P(E_1) = P(E_2) = \frac{1}{2} \)

Bag I contains 4 red and 4 black balls, totaling \( 4+4=8 \) balls.
Probability of drawing a red ball from Bag I: \( P\left(\frac{A}{E_1}\right) = \frac{4}{8} = \frac{1}{2} \)

Bag II contains 2 red and 6 black balls, totaling \( 2+6=8 \) balls.
Probability of drawing a red ball from Bag II: \( P\left(\frac{A}{E_2}\right) = \frac{2}{8} = \frac{1}{4} \)

We need to find the probability that the red ball was drawn from Bag I, given that the drawn ball is red. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{4}} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} \]
For the denominator, find a common denominator (8): \( \frac{1}{4} = \frac{2}{8} \).
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{4}}{\frac{2}{8} + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{1}{4} \cdot \frac{8}{3} = \frac{8}{12} = \frac{2}{3} \]
The probability that the red ball was drawn from Bag I is \( \frac{2}{3} \). This calculation helps identify the source of an observed outcome.
In simple words: We picked a bag, drew a red ball, and want to know if it came from Bag 1. Since Bag 1 has a higher chance of giving a red ball ( \( \frac{1}{2} \) vs \( \frac{1}{4} \) from Bag 2), it's more likely it came from Bag 1. The exact chance is \( \frac{2}{3} \).

🎯 Exam Tip: Remember that "selected at random" for bags means their initial probabilities are equal. Be careful not to confuse \( P(\frac{A}{E_1}) \) with \( P(\frac{E_1}{A}) \).

Question 5. There are three coins. One is having head on both faces, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer: Let \( E_1 \) be the event that the two-headed coin is chosen.
Let \( E_2 \) be the event that the biased coin is chosen.
Let \( E_3 \) be the event that the unbiased coin is chosen.
Let \( A \) be the event that the tossed coin shows heads.

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

Now, let's find the probability of getting heads for each coin:
If the two-headed coin is chosen (event \( E_1 \)), it will always show heads.
\( P\left(\frac{A}{E_1}\right) = 1 \)

If the biased coin is chosen (event \( E_2 \)), it comes up heads 75% of the time.
\( P\left(\frac{A}{E_2}\right) = 75\% = \frac{75}{100} = \frac{3}{4} \)

If the unbiased coin is chosen (event \( E_3 \)), it comes up heads 50% of the time.
\( P\left(\frac{A}{E_3}\right) = \frac{1}{2} \)

We need to find the probability that the two-headed coin was chosen, given that it shows heads. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right) + P(E_3) \cdot P\left(\frac{A}{E_3}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot \frac{1}{2}} \]
Since \( \frac{1}{3} \) is common in all terms in the denominator, it can be factored out and cancelled with the numerator:
\[ P\left(\frac{E_1}{A}\right) = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}} \]
Find a common denominator for the denominator (4):
\[ P\left(\frac{E_1}{A}\right) = \frac{1}{\frac{4}{4} + \frac{3}{4} + \frac{2}{4}} = \frac{1}{\frac{4+3+2}{4}} = \frac{1}{\frac{9}{4}} \]
\[ P\left(\frac{E_1}{A}\right) = 1 \cdot \frac{4}{9} = \frac{4}{9} \]
The probability that the coin chosen was the two-headed coin, given that it showed heads, is \( \frac{4}{9} \). This shows how the observation of heads makes it more likely that a coin designed to show heads often was chosen.
In simple words: We chose one of three coins, flipped it, and got heads. We want to know the chance it was the special coin with heads on both sides. Because that coin always gives heads, and the others only sometimes, the chance is \( \frac{4}{9} \) that it was the two-headed coin.

🎯 Exam Tip: Always remember that a two-headed coin has a 100% probability of showing heads, while an unbiased coin has 50%. These fixed probabilities are crucial for correct calculations.

Question 6. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested. If 0.1 percent of the population actually has the disease. What is the probability that a person has the disease given that his test result is positive?
Answer: Let \( E_1 \) be the event that a person has the disease.
Let \( E_2 \) be the event that a person does not have the disease (is healthy).
Let \( A \) be the event that the test result is positive.

Given probabilities:
0.1% of the population has the disease.
\( P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001 \)

Therefore, the probability that a person does not have the disease:
\( P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0.999 \)

The test is 99% effective in detecting the disease when it is present (true positive rate).
\( P\left(\frac{A}{E_1}\right) = 99\% = \frac{99}{100} = 0.99 \)

The test yields a false positive for 0.5% of healthy people.
\( P\left(\frac{A}{E_2}\right) = 0.5\% = \frac{0.5}{100} = 0.005 \)

We need to find the probability that a person has the disease given that their test result is positive. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{0.001 \cdot 0.99}{0.001 \cdot 0.99 + 0.999 \cdot 0.005} \]
Calculate the products:
\( 0.001 \cdot 0.99 = 0.00099 \)
\( 0.999 \cdot 0.005 = 0.004995 \)

\( \implies \)
\[ P\left(\frac{E_1}{A}\right) = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985} \]
To simplify the fraction, multiply the numerator and denominator by 100,000 (to remove decimals):
\[ P\left(\frac{E_1}{A}\right) = \frac{99}{598.5} = \frac{990}{5985} \]
Both numbers are divisible by 5, then by 9, then by 11:
\( \frac{990}{5985} = \frac{198}{1197} \)
Divide by 9: \( \frac{198 \div 9}{1197 \div 9} = \frac{22}{133} \)
So, the probability is \( \frac{22}{133} \). This shows that even with a positive test, the chance of having the disease is still relatively low if the disease is rare in the population.
In simple words: We want to know the real chance someone has a disease if their test is positive. Even if the test is mostly good, if the disease is very rare, a positive result doesn't always mean you have it for sure. Here, the chance is \( \frac{22}{133} \).

🎯 Exam Tip: Pay close attention to converting percentages to decimals correctly, especially for small percentages like 0.1% or 0.5%. A misplaced decimal can lead to significant errors.

Question 8. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver?
Answer: Let \( E_1 \) be the event that the insured person is a scooter driver.
Let \( E_2 \) be the event that the insured person is a car driver.
Let \( E_3 \) be the event that the insured person is a truck driver.
Let \( A \) be the event that the insured person meets with an accident.

Total number of insured drivers = \( 2000 + 4000 + 6000 = 12000 \).

Probabilities of selecting each type of driver:
\( P(E_1) = \frac{2000}{12000} = \frac{1}{6} \)
\( P(E_2) = \frac{4000}{12000} = \frac{1}{3} \)
\( P(E_3) = \frac{6000}{12000} = \frac{1}{2} \)

Probabilities of an accident for each driver type:
\( P\left(\frac{A}{E_1}\right) = 0.01 \)
\( P\left(\frac{A}{E_2}\right) = 0.03 \)
\( P\left(\frac{A}{E_3}\right) = 0.15 \)

We need to find the probability that the person is a scooter driver, given that they had an accident. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right) + P(E_3) \cdot P\left(\frac{A}{E_3}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{6} \cdot 0.01}{\frac{1}{6} \cdot 0.01 + \frac{1}{3} \cdot 0.03 + \frac{1}{2} \cdot 0.15} \]
Calculate the products in the denominator:
\( \frac{1}{6} \cdot 0.01 = \frac{0.01}{6} \)
\( \frac{1}{3} \cdot 0.03 = \frac{0.03}{3} = 0.01 \)
\( \frac{1}{2} \cdot 0.15 = \frac{0.15}{2} = 0.075 \)

\[ P\left(\frac{E_1}{A}\right) = \frac{0.01/6}{0.01/6 + 0.01 + 0.075} \] Multiply numerator and denominator by 6 to clear the fraction in the first term:
\[ P\left(\frac{E_1}{A}\right) = \frac{0.01}{0.01 + 0.06 + 0.45} = \frac{0.01}{0.52} \]
To remove decimals, multiply numerator and denominator by 100:
\[ P\left(\frac{E_1}{A}\right) = \frac{1}{52} \]
The probability that the person who met with an accident is a scooter driver is \( \frac{1}{52} \). This calculation helps identify which group is most likely involved in an accident given the outcome.
In simple words: An insured person had an accident. We want to find the chance they were a scooter driver. Even though there are many truck and car drivers, and they have higher accident rates, we use Bayes' Theorem to find the exact probability, which is \( \frac{1}{52} \).

🎯 Exam Tip: When using decimals and fractions together, convert all numbers to decimals or all to fractions for easier calculation, usually whichever leads to fewer complex steps.

Question 9. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \( \frac{3}{4} \) be probability that he knows the answer and \( \frac{1}{4} \) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \( \frac{1}{4} \). What is the probability that the student knows the answer given that he answered it correctly?
Answer: Let \( E_1 \) be the event that the student knows the answer.
Let \( E_2 \) be the event that the student guesses the answer.
Let \( A \) be the event that the student answers correctly.

Given probabilities:
Probability that the student knows the answer: \( P(E_1) = \frac{3}{4} \)
Probability that the student guesses the answer: \( P(E_2) = \frac{1}{4} \)

If the student knows the answer, the probability of answering correctly is 1 (certainty).
\( P\left(\frac{A}{E_1}\right) = 1 \)

If the student guesses the answer, the probability of answering correctly is \( \frac{1}{4} \).
\( P\left(\frac{A}{E_2}\right) = \frac{1}{4} \)

We need to find the probability that the student knows the answer given that they answered correctly. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1 + \frac{1}{4} \cdot \frac{1}{4}} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} \]
For the denominator, find a common denominator (16): \( \frac{3}{4} = \frac{12}{16} \).
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{3}{4}}{\frac{12}{16} + \frac{1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{3}{4} \cdot \frac{16}{13} = \frac{3 \cdot 4}{13} = \frac{12}{13} \]
The probability that the student knew the answer, given that they answered correctly, is \( \frac{12}{13} \). This calculation helps understand the likelihood of true knowledge behind a correct answer.
In simple words: If a student gets a multiple-choice question right, what's the chance they actually knew the answer, not just guessed? Since they are more likely to know the answer than guess, and knowing means 100% correct, the chance they knew is very high, specifically \( \frac{12}{13} \).

🎯 Exam Tip: Clearly distinguish between the probability of knowing the answer and the conditional probability of being correct given they know it. When a student knows the answer, the probability of being correct is 1.

Question 10. Suppose 5% men and 0.25% women has white hairs. A person with white hair is chosen at random. Find the probability that a male is selected. Assume that there are equal number of men and women.
Answer: Let \( E_1 \) be the event that the selected person is male.
Let \( E_2 \) be the event that the selected person is female.
Let \( A \) be the event that the selected person has white hair.

Since there are an equal number of men and women in the population,
\( P(E_1) = P(E_2) = \frac{1}{2} \)

Given probabilities of having white hair:
For men: \( P\left(\frac{A}{E_1}\right) = 5\% = \frac{5}{100} = 0.05 \)
For women: \( P\left(\frac{A}{E_2}\right) = 0.25\% = \frac{0.25}{100} = 0.0025 \)

We need to find the probability that the person chosen is male, given that they have white hair. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{2} \cdot 0.05}{\frac{1}{2} \cdot 0.05 + \frac{1}{2} \cdot 0.0025} \]
Since \( \frac{1}{2} \) is common in all terms, it can be factored out and cancelled:
\[ P\left(\frac{E_1}{A}\right) = \frac{0.05}{0.05 + 0.0025} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{0.05}{0.0525} \]
To remove decimals, multiply the numerator and denominator by 10,000:
\[ P\left(\frac{E_1}{A}\right) = \frac{500}{525} \]
Both numbers are divisible by 25:
\( 500 \div 25 = 20 \)
\( 525 \div 25 = 21 \)
So, the probability is \( \frac{20}{21} \). This implies that if a person has white hair, it is highly probable that they are male, because a much larger percentage of men have white hair compared to women.
In simple words: If you pick a person with white hair, what's the chance they are a man? Even though there are equal numbers of men and women, men are much more likely to have white hair. So, if someone has white hair, the chance they are a man is very high, specifically \( \frac{20}{21} \).

🎯 Exam Tip: When given percentages for population groups, convert them to probabilities. If population sizes are equal, the prior probabilities \( P(E_1) \) and \( P(E_2) \) will be \( \frac{1}{2} \).

Question 11. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer: Let \( E_1 \) be the event that the first group wins.
Let \( E_2 \) be the event that the second group wins.
Let \( A \) be the event that a new product is introduced.

Given probabilities of winning:
\( P(E_1) = 0.6 \)
\( P(E_2) = 0.4 \)

Given probabilities of introducing a new product:
If the first group wins: \( P\left(\frac{A}{E_1}\right) = 0.7 \)
If the second group wins: \( P\left(\frac{A}{E_2}\right) = 0.3 \)

We need to find the probability that the new product was introduced by the second group, given that a new product was introduced. This is \( P\left(\frac{E_2}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_2}{A}\right) = \frac{0.4 \cdot 0.3}{0.6 \cdot 0.7 + 0.4 \cdot 0.3} \]
Calculate the products:
\( 0.4 \cdot 0.3 = 0.12 \)
\( 0.6 \cdot 0.7 = 0.42 \)

\( \implies \)
\[ P\left(\frac{E_2}{A}\right) = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} \]
To remove decimals, multiply numerator and denominator by 100:
\[ P\left(\frac{E_2}{A}\right) = \frac{12}{54} \]
Both numbers are divisible by 6:
\( 12 \div 6 = 2 \)
\( 54 \div 6 = 9 \)
So, the probability is \( \frac{2}{9} \). This problem shows how conditional probability can be used in business scenarios to attribute outcomes to specific groups.
In simple words: A new product was launched. We want to know the chance it was introduced by the second group. We know the chance each group wins and their chance of launching a new product. Combining these, the probability that the second group introduced it is \( \frac{2}{9} \).

🎯 Exam Tip: Ensure that all probabilities are correctly converted from percentages or raw numbers to decimals (if not already) before performing calculations. Double-check your multiplication and addition steps for accuracy.

Question 12. Suppose a girl throws a dice. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 then tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die ?
Answer: Let \( E_1 \) be the event that the girl throws a 5 or 6 on the die.
Let \( E_2 \) be the event that the girl throws a 1, 2, 3, or 4 on the die.
Let \( A \) be the event that she obtains exactly one head.

There are 6 possible outcomes when throwing a die: {1, 2, 3, 4, 5, 6}.
Probability of throwing a 5 or 6: \( P(E_1) = \frac{2}{6} = \frac{1}{3} \)
Probability of throwing a 1, 2, 3, or 4: \( P(E_2) = \frac{4}{6} = \frac{2}{3} \)

Now, let's find the probability of obtaining exactly one head in each scenario:
Case 1: If she throws a 5 or 6 (event \( E_1 \)), she tosses a coin three times. The possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (8 outcomes). Exactly one head occurs for HTT, THT, TTH (3 outcomes).
So, \( P\left(\frac{A}{E_1}\right) = \frac{3}{8} \)

Case 2: If she throws a 1, 2, 3, or 4 (event \( E_2 \)), she tosses a coin once. The possible outcomes are H, T (2 outcomes). Exactly one head occurs for H (1 outcome).
So, \( P\left(\frac{A}{E_2}\right) = \frac{1}{2} \)

We need to find the probability that she threw 1, 2, 3, or 4 (event \( E_2 \)), given that she obtained exactly one head (event \( A \)). This is \( P\left(\frac{E_2}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_2}{A}\right) = \frac{P(E_2) \cdot P\left(\frac{A}{E_2}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8} + \frac{2}{3} \cdot \frac{1}{2}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{2}{6}}{\frac{3}{24} + \frac{2}{6}} \]
Simplify fractions: \( \frac{2}{6} = \frac{1}{3} \), \( \frac{3}{24} = \frac{1}{8} \).
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} \]
For the denominator, find a common denominator (24):
\( \frac{1}{8} = \frac{3}{24} \)
\( \frac{1}{3} = \frac{8}{24} \)

\( \implies \)
\[ P\left(\frac{E_2}{A}\right) = \frac{\frac{1}{3}}{\frac{3}{24} + \frac{8}{24}} = \frac{\frac{1}{3}}{\frac{11}{24}} \]
\[ P\left(\frac{E_2}{A}\right) = \frac{1}{3} \cdot \frac{24}{11} = \frac{8}{11} \]
The probability that she threw 1, 2, 3, or 4, given that she obtained exactly one head, is \( \frac{8}{11} \). This problem elegantly combines dice rolls and coin tosses.
In simple words: A girl rolls a die and then flips coins based on the result. If she ends up with exactly one head, we want to know the chance she rolled a 1, 2, 3, or 4. We calculate the likelihood of getting one head in both scenarios (rolling 5/6 vs. rolling 1-4) and use Bayes' Theorem to find the answer: \( \frac{8}{11} \).

🎯 Exam Tip: When dealing with sequential events like dice rolls followed by coin tosses, clearly define the outcomes and probabilities for each stage. Drawing a small tree diagram can help visualize the possibilities and avoid missing any combinations.

Question 13. A card from a pack of 52 cards is lost. From remaining cards of the pack, two cards are drawn and found to be both diamonds. Find probability of the lost card being a diamond.
Answer: Let \( E_1 \) be the event that the lost card is a diamond.
Let \( E_2 \) be the event that the lost card is not a diamond.
Let \( A \) be the event that two cards drawn from the remaining pack are both diamonds.

In a standard pack of 52 cards, there are 13 diamonds.
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 probability of drawing two diamonds from the remaining 51 cards, given each event:
Case 1: The lost card is a diamond (event \( E_1 \)).
The remaining pack has 51 cards, with 12 diamonds and 39 non-diamonds. The number of ways to draw 2 diamonds from 12 is \( ^{12}C_2 \). The total number of ways to draw 2 cards from 51 is \( ^{51}C_2 \).
\( P\left(\frac{A}{E_1}\right) = \frac{^{12}C_2}{^{51}C_2} = \frac{\frac{12 \cdot 11}{2 \cdot 1}}{\frac{51 \cdot 50}{2 \cdot 1}} = \frac{12 \cdot 11}{51 \cdot 50} = \frac{132}{2550} \)

Case 2: The lost card is not a diamond (event \( E_2 \)).
The remaining pack has 51 cards, with 13 diamonds and 38 non-diamonds. The number of ways to draw 2 diamonds from 13 is \( ^{13}C_2 \). The total number of ways to draw 2 cards from 51 is \( ^{51}C_2 \).
\( P\left(\frac{A}{E_2}\right) = \frac{^{13}C_2}{^{51}C_2} = \frac{\frac{13 \cdot 12}{2 \cdot 1}}{\frac{51 \cdot 50}{2 \cdot 1}} = \frac{13 \cdot 12}{51 \cdot 50} = \frac{156}{2550} \)

We need to find the probability that the lost card was a diamond, given that two drawn cards are diamonds. This is \( P\left(\frac{E_1}{A}\right) \). Using Bayes' Theorem:
\[ P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} \]
Substitute the values:
\[ P\left(\frac{E_1}{A}\right) = \frac{\frac{1}{4} \cdot \frac{132}{2550}}{\frac{1}{4} \cdot \frac{132}{2550} + \frac{3}{4} \cdot \frac{156}{2550}} \]
Notice that \( \frac{1}{4} \) and \( \frac{1}{2550} \) are common factors in all terms. They can be cancelled out.
\[ P\left(\frac{E_1}{A}\right) = \frac{1 \cdot 132}{1 \cdot 132 + 3 \cdot 156} \]
\[ P\left(\frac{E_1}{A}\right) = \frac{132}{132 + 468} = \frac{132}{600} \]
Simplify the fraction by dividing by common factors. Both are divisible by 4:
\( 132 \div 4 = 33 \)
\( 600 \div 4 = 150 \)
\[ P\left(\frac{E_1}{A}\right) = \frac{33}{150} \]
Both are divisible by 3:
\( 33 \div 3 = 11 \)
\( 150 \div 3 = 50 \)
So, the probability is \( \frac{11}{50} \). This result shows how observing the drawn cards changes our belief about the lost card.
In simple words: One card is lost from a deck. Then, two cards are drawn and both are diamonds. We want to find the chance that the *lost* card was also a diamond. By seeing two diamonds drawn from the remaining cards, it becomes less likely that a diamond was lost, but we use calculations to find the exact probability, which is \( \frac{11}{50} \).

🎯 Exam Tip: For problems involving combinations, clearly define the number of items available for selection and the number of items being selected. Be careful to adjust the pool of cards (total and specific types) after a card is lost.

Question 13. A card from a pack of 52 card is lost From remaining cards of the pack, two cards are drawn found to be both diamonds. Find probability of the lost card being a diamond.
Answer: Let \( E_1 \) be the event that the lost card is a diamond. Let \( E_2 \) be the event that the lost card is not a diamond. The total number of cards in a pack is 52. There are 13 diamond cards.
So, the probability that a diamond card is lost is:
\( P(E_1) = \frac{13}{52} = \frac{1}{4} \)
The probability that a non-diamond card is lost is:
\( P(E_2) = 1 - P(E_1) = 1 - \frac{1}{4} = \frac{3}{4} \)
Let \( A \) be the event that two cards drawn from the remaining 51 cards are both diamonds. We need to find the probability that the lost card was a diamond, given that two diamonds were drawn, i.e., \( P(E_1|A) \).
First, calculate the probability of drawing two diamonds if a diamond card was lost:
\( P(A|E_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{12 \times 11}{51 \times 50} \)
Next, calculate the probability of drawing two diamonds if a non-diamond card was lost:
\( P(A|E_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{13 \times 12}{51 \times 50} \)
Now, we use Bayes' theorem to find \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(E_1) \times P(A|E_1)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)} \)
\( \implies P(E_1|A) = \frac{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}}{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50} + \frac{3}{4} \times \frac{13 \times 12}{51 \times 50}} \)
We can simplify this expression by cancelling the common term \( \frac{1}{4 \times 51 \times 50} \) from both the numerator and the denominator. This makes the calculation easier.
\( \implies P(E_1|A) = \frac{12 \times 11}{12 \times 11 + 3 \times 13 \times 12} \)
\( \implies P(E_1|A) = \frac{132}{132 + 468} \)
\( \implies P(E_1|A) = \frac{132}{600} \)
\( \implies P(E_1|A) = \frac{33}{150} \) (dividing by 4)
\( \implies P(E_1|A) = \frac{11}{50} \) (dividing by 3)
Therefore, the probability that the lost card was a diamond is \( \frac{11}{50} \).
In simple words: We calculate the chances of losing a diamond card versus a non-diamond card. Then, we find the chance of drawing two diamonds in both situations. Finally, we use Bayes' theorem to find the probability that the lost card was a diamond, given that we actually drew two diamonds.

🎯 Exam Tip: When using Bayes' theorem for 'lost item' problems, clearly define the events for the item being lost and for the observation made. Remember that the total number of items changes after one is lost.

Question 14. A bag contains 3 red and 7 black. Two balls are drawn one by one at a time at random without replacement. If second drawn ball is red, then what is the probability that first drawn ball is also red?
Answer: Let \( E_1 \) be the event that the first ball drawn is red. Let \( E_2 \) be the event that the first ball drawn is black. Let \( B \) be the event that the second ball drawn is red. We want to find the probability that the first ball drawn was red, given that the second ball drawn is red, which is \( P(E_1|B) \).
Total balls in the bag = 3 Red + 7 Black = 10 balls.
First, we find the probabilities of the first ball being red or black:
\( P(E_1) = \text{Probability of first ball being red} = \frac{3}{10} \)
\( P(E_2) = \text{Probability of first ball being black} = \frac{7}{10} \)
Next, we find the conditional probabilities of the second ball being red:
\( P(B|E_1) = \text{Probability of second ball being red, given first was red} \)
If the first ball was red, 2 red balls and 7 black balls remain (total 9 balls).
\( P(B|E_1) = \frac{2}{9} \)
\( P(B|E_2) = \text{Probability of second ball being red, given first was black} \)
If the first ball was black, 3 red balls and 6 black balls remain (total 9 balls).
\( P(B|E_2) = \frac{3}{9} = \frac{1}{3} \)
Now, we find the total probability of the second ball being red, \( P(B) \), using the law of total probability:
\( P(B) = P(B|E_1)P(E_1) + P(B|E_2)P(E_2) \)
\( \implies P(B) = \left(\frac{2}{9}\right) \times \left(\frac{3}{10}\right) + \left(\frac{3}{9}\right) \times \left(\frac{7}{10}\right) \)
\( \implies P(B) = \frac{6}{90} + \frac{21}{90} \)
\( \implies P(B) = \frac{27}{90} = \frac{3}{10} \)
Finally, we use Bayes' theorem to find \( P(E_1|B) \):
\( P(E_1|B) = \frac{P(B|E_1)P(E_1)}{P(B)} \)
\( \implies P(E_1|B) = \frac{\left(\frac{2}{9}\right) \times \left(\frac{3}{10}\right)}{\frac{3}{10}} \)
\( \implies P(E_1|B) = \frac{\frac{6}{90}}{\frac{3}{10}} \)
\( \implies P(E_1|B) = \frac{6}{90} \times \frac{10}{3} \)
\( \implies P(E_1|B) = \frac{60}{270} = \frac{2}{9} \)
The probability that the first ball drawn was red, given the second was red, is \( \frac{2}{9} \). Drawing without replacement affects the probabilities for subsequent draws.
In simple words: We want to know the chance the first ball was red, given the second ball was red. We find the chance of drawing a red first, then another red. We also find the chance of drawing a black first, then a red. We combine these to find the overall chance of the second ball being red. Then, we use these to find our answer.

🎯 Exam Tip: Always clearly define your events \( E_1, E_2, B \) etc. and draw a tree diagram for 'without replacement' problems to visualize the probabilities. This helps in correctly applying the conditional probability formulas and Bayes' theorem.