GSEB Class 12 Maths Solutions Chapter 13 Probability Exercise 13.3

Question 1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned in to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer: The urn initially holds 5 red balls and 5 black balls.
(i) Let's assume a red ball is drawn first.
The probability of drawing a red ball is \( \frac{5}{10} = \frac{1}{2} \).
Now, two red balls are added to the urn.
This means the urn now contains 7 red balls and 5 black balls.
The probability of drawing a red ball in this case is \( \frac{7}{12} \).
(ii) Let's assume a black ball is drawn at the initial attempt.
The probability of drawing a black ball is \( \frac{5}{10} = \frac{1}{2} \).
Next, 2 black balls are put into the urn.
Therefore, the urn now has 5 red balls and 7 black balls.
The probability of drawing a red ball in this case is \( \frac{5}{12} \).
The total probability of drawing a second ball as red is calculated as:
\( P(\text{second ball is red}) = P(\text{red first}) \times P(\text{red second | red first}) + P(\text{black first}) \times P(\text{red second | black first}) \)
\( = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} \)
\( = \frac{7}{24} + \frac{5}{24} \)
\( = \frac{7+5}{24} \)
\( = \frac{12}{24} \)
\( = \frac{1}{2} \)
In simple words: The chance of picking a red ball second depends on what color ball was picked first and then put back with two more of its kind. We look at both possibilities and combine their probabilities to find the overall chance.

Exam Tip: For conditional probability problems involving replacement and addition, break the problem into distinct cases based on the first event. Calculate the probability for each case and then combine them using the law of total probability.

Question 2. 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 \( E_1 \) represent the event of selecting the first bag, and \( E_2 \) represent the event of selecting the second bag.
The probability of choosing either bag is \( \frac{1}{2} \). So, \( P(E_1) = P(E_2) = \frac{1}{2} \).
Let \( A \) be the event that a red ball is drawn.
The first bag has 4 red and 4 black balls (total 8 balls).
The probability of selecting a red ball from the first bag is \( P(A/E_1) = \frac{4}{8} = \frac{1}{2} \).
The second bag has 2 red and 6 black balls (total 8 balls).
The probability of selecting a red ball from the second bag is \( P(A/E_2) = \frac{2}{8} = \frac{1}{4} \).
We need to find the probability that the ball was drawn from the first bag, given that it is red, i.e., \( P(E_1/A) \). We use Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}} \]
\[ = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} \]
\[ = \frac{\frac{1}{4}}{\frac{2+1}{8}} \]
\[ = \frac{\frac{1}{4}}{\frac{3}{8}} \]
\[ = \frac{1}{4} \times \frac{8}{3} \]
\[ = \frac{2}{3} \]
In simple words: We have two bags. We picked one at random and drew a red ball. We want to know the chance that the red ball came from the first bag. We use a formula that considers the probability of picking each bag and the probability of drawing a red ball from each bag.

Exam Tip: Bayes' Theorem is crucial for problems where you need to find the probability of a cause given an observed effect. Clearly define events and identify all conditional probabilities before applying the formula.

Question 3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previouis year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholar attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is hosteler?
Answer: Let \( E_1 \) represent the event that a student lives in the hostel, and \( E_2 \) represent the event that a student is a day scholar.
The probability of a student being a hosteler is \( P(E_1) = 60\% = 0.6 \).
The probability of a student being a day scholar is \( P(E_2) = 40\% = 0.4 \).
Let \( A \) be the event that a student receives an A grade.
The probability that a hosteler gets an A grade is \( P(A/E_1) = 30\% = 0.3 \).
The probability that a day scholar gets an A grade is \( P(A/E_2) = 20\% = 0.2 \).
We need to find the probability that a student is a hosteler, given that they received an A grade, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} \]
\[ = \frac{0.18}{0.18 + 0.08} \]
\[ = \frac{0.18}{0.26} \]
\[ = \frac{18}{26} \]
\[ = \frac{9}{13} \]
In simple words: We know the chances of students living in a hostel or being day scholars, and how likely each group is to get an A grade. If we pick a student with an A grade, we want to find the chance that this student lives in the hostel.

Exam Tip: Convert all percentages to decimals when performing calculations. Clearly label your events and probabilities to avoid confusion, especially in problems involving multiple conditions.

Question 4. In answering a question on a multiple choice test, a student knows the answer or guesses. Let \( \frac{3}{4} \) be the 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 a student who knows the answer, given that he answered it correctly?
Answer: Let \( E_1 \) be the event that the student knows the answer, and \( E_2 \) be the event that the student guesses the answer.
The probability that the student knows the answer is \( P(E_1) = \frac{3}{4} \).
The probability that the student guesses the answer is \( P(E_2) = \frac{1}{4} \).
Let \( A \) be the event that the answer is correct.
If the student knows the answer, then they will answer correctly with probability \( P(A/E_1) = 1 \).
If the student guesses the answer, they will answer correctly with probability \( P(A/E_2) = \frac{1}{4} \).
We need to find the probability that the student knew the answer, given that they answered correctly, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} \]
\[ = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} \]
\[ = \frac{\frac{3}{4}}{\frac{12}{16} + \frac{1}{16}} \]
\[ = \frac{\frac{3}{4}}{\frac{13}{16}} \]
\[ = \frac{3}{4} \times \frac{16}{13} \]
\[ = \frac{12}{13} \]
In simple words: A student either knows the answer or guesses. If they guess, they have a small chance of being right. If we know the student answered correctly, we want to find the likelihood that they actually knew the answer.

Exam Tip: Remember that "knows the answer" implies a 100% chance of being correct (\( P(A/E_1) = 1 \)) unless specified otherwise. This is a common point of confusion.

Question 5. A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). 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 \( E \) represent the event that a person has the disease, and \( E' \) represent the event that a person does not have the disease.
The probability that a person has the disease is \( P(E) = 0.1\% = 0.001 \).
The probability that a person does not have the disease is \( P(E') = 1 - P(E) = 1 - 0.001 = 0.999 \).
Let \( A \) be the event that the blood test result is positive.
The probability of a positive test given the person has the disease (true positive) is \( P(A/E) = 99\% = 0.99 \).
The probability of a positive test given the person does not have the disease (false positive) is \( P(A/E') = 0.5\% = 0.005 \).
We need to find the probability that a person has the disease given a positive test result, i.e., \( P(E/A) \). Using Bayes' Theorem:
\[ P(E/A) = \frac{P(E)P(A/E)}{P(E)P(A/E) + P(E')P(A/E')} \]
\[ = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} \]
\[ = \frac{0.00099}{0.00099 + 0.004995} \]
\[ = \frac{0.00099}{0.005985} \]
\[ = \frac{990}{5985} \]
\[ = \frac{198}{1197} \]
\[ = \frac{22}{133} \]
In simple words: We want to know the chance that someone actually has a disease if their test result is positive. This calculation considers how common the disease is, how good the test is at finding it, and how often the test gives a wrong positive result for healthy people.

Exam Tip: Be very careful converting percentages to decimals, especially for small percentages like 0.1% or 0.5%. A common mistake is to write 0.1% as 0.1 instead of 0.001. This is a classic Bayes' theorem problem where the "false positive" rate can significantly affect the overall probability.

Question 6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up head 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed. It shows head. What is the probability that it was the two headed coin?
Answer: Let \( E_1 \), \( E_2 \), and \( E_3 \) be the events of selecting the two-headed coin, the biased coin, and the unbiased coin, respectively.
Since one of the three coins is chosen at random, the probability of selecting any coin is \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \).
Let \( A \) be the event that the tossed coin shows a head.
For the two-headed coin, it will always show head. So, \( P(A/E_1) = 1 \).
For the biased coin, head comes up 75% of the time. So, \( P(A/E_2) = 0.75 = \frac{3}{4} \).
For the unbiased coin, head comes up 50% of the time. So, \( P(A/E_3) = \frac{1}{2} \).
We need to find the probability that it was the two-headed coin given that a head was shown, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \]
\[ = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}} \]
\[ = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{4} + \frac{1}{6}} \]
To combine the denominators, find a common denominator, which is 12:
\[ = \frac{\frac{1}{3}}{\frac{4}{12} + \frac{3}{12} + \frac{2}{12}} \]
\[ = \frac{\frac{1}{3}}{\frac{4+3+2}{12}} \]
\[ = \frac{\frac{1}{3}}{\frac{9}{12}} \]
\[ = \frac{\frac{1}{3}}{\frac{3}{4}} \]
\[ = \frac{1}{3} \times \frac{4}{3} \]
\[ = \frac{4}{9} \]
In simple words: We have three different coins: one with two heads, one that usually lands on heads, and one that is fair. If we randomly pick one, toss it, and it lands on heads, we want to calculate the chance that we picked the special two-headed coin.

Exam Tip: When dealing with multiple mutually exclusive initial events (like selecting different coins), remember to sum their weighted probabilities in the denominator of Bayes' Theorem. Always simplify fractions at each step to make calculations easier.

Question 7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident by them 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: The total number of drivers insured is \( 2000 + 4000 + 6000 = 12000 \).
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.
The probabilities of selecting each type of driver are:
\( P(E_1) = \frac{2000}{12000} = \frac{1}{6} \)
\( P(E_2) = \frac{4000}{12000} = \frac{1}{3} \)
\( P(E_3) = \frac{6000}{12000} = \frac{1}{2} \)
Let \( A \) be the event that the insured person meets with an accident.
The probability of an accident for a scooter driver is \( P(A/E_1) = 0.01 \).
The probability of an accident for a car driver is \( P(A/E_2) = 0.03 \).
The probability of an accident for a truck driver is \( P(A/E_3) = 0.15 \).
We need to find the probability that the person is a scooter driver, given that they met with an accident, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \]
\[ = \frac{\frac{1}{6} \times 0.01}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15} \]
\[ = \frac{0.01/6}{0.01/6 + 0.03/3 + 0.15/2} \]
\[ = \frac{0.01}{0.01 + 2 \times 0.03 + 3 \times 0.15} \]
\[ = \frac{0.01}{0.01 + 0.06 + 0.45} \]
\[ = \frac{0.01}{0.52} \]
\[ = \frac{1}{52} \]
In simple words: An insurance company covers many types of drivers, each with a different chance of having an accident. If someone gets into an accident, we want to figure out the chance that this person was a scooter driver.

Exam Tip: Be meticulous with calculations involving decimals and fractions. It's often helpful to keep values as fractions until the final step or convert everything to decimals early on to ensure consistency.

Question 8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random rom this and is found to be defective. What is the probability that it was produced by machine B?
Answer: Let \( E_1 \) be the event that an item is produced by machine A, and \( E_2 \) be the event that an item is produced by machine B.
The probability that an item is produced by machine A is \( P(E_1) = 60\% = 0.6 \).
The probability that an item is produced by machine B is \( P(E_2) = 40\% = 0.4 \).
Let \( A \) be the event that a selected item is defective.
The probability that an item produced by machine A is defective is \( P(A/E_1) = 2\% = 0.02 \).
The probability that an item produced by machine B is defective is \( P(A/E_2) = 1\% = 0.01 \).
We need to find the probability that a defective item was produced by machine B, i.e., \( P(E_2/A) \). Using Bayes' Theorem:
\[ P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} \]
\[ = \frac{0.004}{0.012 + 0.004} \]
\[ = \frac{0.004}{0.016} \]
\[ = \frac{4}{16} \]
\[ = \frac{1}{4} \]
In simple words: A factory uses two machines, A and B, to make items. We know how much each machine makes and how often their items are faulty. If we pick a faulty item, we want to find the likelihood it came from machine B.

Exam Tip: When dealing with production and defect rates, always ensure you convert percentages to decimals correctly. Carefully distinguish between the probability of a machine producing an item and the conditional probability of that item being defective.

Question 9. Two groups are competing for the positions on the board of directors of a corporation. The probabilities that the first and 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 was introduced by the second group.
Answer: Let \( E_1 \) be the event that the first group wins, and \( E_2 \) be the event that the second group wins.
The probability that the first group wins is \( P(E_1) = 0.6 \).
The probability that the second group wins is \( P(E_2) = 0.4 \).
Let \( A \) be the event that a new product is introduced.
The probability of a new product being introduced if the first group wins is \( P(A/E_1) = 0.7 \).
The probability of a new product being introduced if the second group wins is \( P(A/E_2) = 0.3 \).
We need to find the probability that the new product was introduced by the second group, given that a new product was introduced, i.e., \( P(E_2/A) \). Using Bayes' Theorem:
\[ P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} \]
\[ = \frac{0.12}{0.42 + 0.12} \]
\[ = \frac{0.12}{0.54} \]
\[ = \frac{12}{54} \]
\[ = \frac{2}{9} \]
In simple words: We know the chances of two groups winning an election and how likely each group is to launch a new product if they win. If a new product is launched, we want to figure out the chance that the second group was the one who won.

Exam Tip: Pay close attention to the phrasing of the question to determine which conditional probability you need to calculate (e.g., \( P(E_1/A) \) vs. \( P(E_2/A) \)). This is vital for correctly setting up Bayes' Theorem.

Question 10. 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, she 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, 2, 3 or 4 with the dice?
Answer: When a die is thrown, there are 6 possible outcomes (1, 2, 3, 4, 5, 6).
Let \( E_1 \) be the event that she gets a 5 or 6 on the die.
The probability of \( E_1 \) is \( P(E_1) = \frac{2}{6} = \frac{1}{3} \). In this case, she tosses a coin three times.
Let \( E_2 \) be the event that she gets a 1, 2, 3, or 4 on the die.
The probability of \( E_2 \) is \( P(E_2) = \frac{4}{6} = \frac{2}{3} \). In this case, she tosses a coin once.
Let \( A \) be the event that she obtains exactly one head.
If \( E_1 \) occurs (die roll 5 or 6), she tosses a coin three times. The possible outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. There are \( 2^3 = 8 \) total outcomes.
The outcomes with exactly one head are {HTT, THT, TTH}. There are 3 such outcomes.
So, the probability of getting exactly one head given \( E_1 \) is \( P(A/E_1) = \frac{3}{8} \).
If \( E_2 \) occurs (die roll 1, 2, 3, or 4), she tosses a coin once. The possible outcomes are {H, T}. There are \( 2^1 = 2 \) total outcomes.
The outcomes with exactly one head is {H}. There is 1 such outcome.
So, the probability of getting exactly one head given \( E_2 \) is \( P(A/E_2) = \frac{1}{2} \).
We need to find the probability that she threw 2, 3, or 4 (i.e., event \( E_2 \)) given that she obtained exactly one head, i.e., \( P(E_2/A) \). Using Bayes' Theorem:
\[ P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} \]
\[ = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} \]
To combine the denominators, find a common denominator, which is 24:
\[ = \frac{\frac{1}{3}}{\frac{3}{24} + \frac{8}{24}} \]
\[ = \frac{\frac{1}{3}}{\frac{11}{24}} \]
\[ = \frac{1}{3} \times \frac{24}{11} \]
\[ = \frac{8}{11} \]
In simple words: A girl rolls a die, and depending on the number, she flips a coin either once or three times. If she ends up with exactly one head, we want to know the chance that her initial die roll was a 1, 2, 3, or 4.

Exam Tip: For problems involving sequential events (like rolling a die, then tossing coins), clearly define the outcomes and probabilities for each stage. Listing sample spaces for coin tosses helps avoid errors in calculating conditional probabilities like \( P(A/E_1) \).

Question 11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of time, B on the. job for 30% of the time and C is on the job for 20% of the time. If a defective item is produced, what is the probability that it was produced by A?
Answer: Let \( E_1 \), \( E_2 \), and \( E_3 \) be the events that the item is produced by operator A, B, and C, respectively.
The time spent by each operator is:
\( P(E_1) = 50\% = 0.5 \)
\( P(E_2) = 30\% = 0.3 \)
\( P(E_3) = 20\% = 0.2 \)
Let \( A \) be the event that a defective item is produced.
The probability of a defective item given it was produced by operator A is \( P(A/E_1) = 1\% = 0.01 \).
The probability of a defective item given it was produced by operator B is \( P(A/E_2) = 5\% = 0.05 \).
The probability of a defective item given it was produced by operator C is \( P(A/E_3) = 7\% = 0.07 \).
We need to find the probability that the defective item was produced by operator A, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)} \]
\[ = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} \]
\[ = \frac{0.005}{0.005 + 0.015 + 0.014} \]
\[ = \frac{0.005}{0.034} \]
\[ = \frac{5}{34} \]
In simple words: Three workers make items, and each worker has a different rate of making faulty products. They also work for different amounts of time. If we find a faulty item, we want to know the chance it was made by operator A.

Exam Tip: This type of problem often involves calculating the "prior" probabilities (time on job) and "conditional" probabilities (defect rates). Make sure to correctly multiply these in the numerator and sum all possibilities in the denominator according to Bayes' theorem.

Question 12. A card from a pack of 52 cards is lost. From the remaining cards of the pack; two cards are drawn and found to be both diamond. Find the 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.
There are 13 diamond cards in a pack of 52 cards.
The probability that the lost card is a diamond is \( P(E_1) = \frac{13}{52} = \frac{1}{4} \).
The probability that the lost card is not a diamond 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.
(i) If the lost card is a diamond (\( E_1 \) occurs), then there are 12 diamond cards and 39 non-diamond cards left in the remaining 51 cards.
The probability of drawing two diamonds from these 51 cards is \( P(A/E_1) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{\frac{12 \times 11}{2}}{\frac{51 \times 50}{2}} = \frac{12 \times 11}{51 \times 50} \).
(ii) If the lost card is not a diamond (\( E_2 \) occurs), then there are 13 diamond cards and 38 non-diamond cards left in the remaining 51 cards.
The probability of drawing two diamonds from these 51 cards is \( P(A/E_2) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{\frac{13 \times 12}{2}}{\frac{51 \times 50}{2}} = \frac{13 \times 12}{51 \times 50} \).
We need to find the probability that the lost card was a diamond, given that two drawn cards are diamonds, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}}{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50} + \frac{3}{4} \times \frac{13 \times 12}{51 \times 50}} \]
We can cancel \( \frac{1}{4 \times 51 \times 50} \) from all terms:
\[ = \frac{12 \times 11}{12 \times 11 + 3 \times 13 \times 12} \]
\[ = \frac{132}{132 + 468} \]
\[ = \frac{132}{600} \]
\[ = \frac{33}{150} \]
\[ = \frac{11}{50} \]
In simple words: A card is lost from a deck. We then pick two cards from the remaining ones, and both happen to be diamonds. We want to find the chance that the card we lost was also a diamond.

Exam Tip: Problems involving combinations (\( \binom{n}{k} \)) often appear in probability questions. Remember to carefully calculate these combinations for each scenario. Look for common factors in the numerator and denominator to simplify calculations, as shown in this solution.

Question 13. Probability that A speaks truth is \( \frac{4}{5} \). A reports that head appears. The probability that actually there was head is:
(A) \( \frac{4}{5} \)
(B) \( \frac{1}{2} \)
(C) \( \frac{1}{5} \)
(D) \( \frac{2}{5} \)
Answer: (A) \( \frac{4}{5} \)
Let \( E_1 \) be the event that A speaks the truth.
Let \( E_2 \) be the event that A does not speak the truth.
The probability that A speaks the truth is \( P(E_1) = \frac{4}{5} \).
The probability that A does not speak the truth is \( P(E_2) = 1 - P(E_1) = 1 - \frac{4}{5} = \frac{1}{5} \).
Let \( H \) be the event that a head actually appears.
Let \( A \) be the event that A reports a head appears.
If A speaks the truth (\( E_1 \)), then A reports a head, and a head actually appeared. The probability of getting a head on a coin toss is \( \frac{1}{2} \). So, \( P(A/E_1) = \frac{1}{2} \).
If A does not speak the truth (\( E_2 \)), then A reports a head, but a head did not actually appear (meaning a tail appeared). The probability of getting a tail on a coin toss is \( \frac{1}{2} \). So, \( P(A/E_2) = \frac{1}{2} \).
We need to find the probability that a head actually appeared, given that A reports a head, i.e., \( P(E_1/A) \). Using Bayes' Theorem:
\[ P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)} \]
\[ = \frac{\frac{4}{5} \times \frac{1}{2}}{\frac{4}{5} \times \frac{1}{2} + \frac{1}{5} \times \frac{1}{2}} \]
\[ = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{1}{10}} \]
\[ = \frac{\frac{4}{10}}{\frac{5}{10}} \]
\[ = \frac{4}{5} \]
In simple words: We know how often someone tells the truth. They say a coin landed on heads. We want to find the real chance that it actually was heads, considering if they are truthful or not.

Exam Tip: In "truth-teller" problems, "reporting an event" can happen in two ways: either the person tells the truth and the event occurred, or the person lies and the opposite event occurred. Carefully define these conditions for your conditional probabilities.

Question 14. If A and B are two events such that A \( \subset \) B and P(B) \( \neq \) 0, then which of the following is true?
(A) P(A/B) = \( \frac{P(B)}{P(A)} \)
(B) P(A/B) < P(A)
(C) P(A/B) \( \geq \) P(A)
(D) None of the options
Answer: (C) P(A/B) \( \geq \) P(A)
Given that A \( \subset \) B, this means that if event A occurs, event B must also occur. Therefore, the intersection of A and B is simply A, i.e., \( A \cap B = A \).
The probability of the intersection is \( P(A \cap B) = P(A) \).
The conditional probability of A given B is defined as:
\[ P(A/B) = \frac{P(A \cap B)}{P(B)} \]
Since \( A \cap B = A \), we can substitute \( P(A \cap B) \) with \( P(A) \):
\[ P(A/B) = \frac{P(A)}{P(B)} \]
We know that for any event B, \( P(B) \leq 1 \).
Since \( P(B) \leq 1 \) and \( P(B) \neq 0 \), dividing \( P(A) \) by a number less than or equal to 1 will make the result greater than or equal to \( P(A) \).
So, \( P(A/B) \geq P(A) \).
Therefore, option (C) is the correct answer.
In simple words: If event A is completely inside event B (like a smaller circle inside a bigger circle), then the chance of A happening when B has already happened is at least as big as the chance of A happening normally. This is because knowing B happened makes A more likely, as A can only happen if B also happens.

Exam Tip: Understanding set relationships like A \( \subset \) B (A is a subset of B) is crucial for simplifying probability expressions. Remember that when A is a subset of B, the occurrence of B makes A more likely or at least as likely, hence \( P(A/B) \geq P(A) \).