Welcome! If you’re looking for the NCERT Solutions for Class 12 Maths Chapter 13, Exercise 13.3, you’re in the right place. Below, you will find detailed, step-by-step solutions for every question in Exercise 13.3, designed to help you clear all your doubts and understand the logic behind each answer.
This exercise introduces the core concepts of Baye’s Theorem, which are crucial for your board exams.
Q1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned 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?
Solution:
We need to find the total probability of the second ball being red, which can happen under two different conditions: the first ball was red, or the first ball was black.
Step 1: Define the events
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 A be the event that the second ball drawn is Red.
We want to find P(A).
Step 2: Find the probabilities of the first draw
The urn initially has 5 Red and 5 Black balls (Total 10).
The probability of drawing a Red ball first is:
P(E_1) = \frac{5}{10} = \frac{1}{2}
The probability of drawing a Black ball first is:
P(E_2) = \frac{5}{10} = \frac{1}{2}
Step 3: Finding conditional probabilities for the second draw
We need to find the probability of A (second ball is Red) given that E_1 or E_2 has occurred.
Case 1: Probability of A given E_1 occurred (P(A|E_1))
If the first ball was Red (E_1), it is returned, and 2 more Red balls are added.
The urn now contains: (5 + 2) Red and 5 Black = 7 Red and 5 Black.
The total number of balls is 7 + 5 = 12 .
So, P(A|E_1) = \frac{7}{12}
Case 2: Probability of A given E_2 occurred (P(A|E_2))
If the first ball was Black (E_2), it is returned, and 2 more Black balls are added.
The urn now contains: 5 Red and (5 + 2) Black = 5 Red and 7 Black.
The total number of balls is 5 + 7 = 12 .
So, P(A|E_2) = \frac{5}{12}
Step 4: Apply the Theorem of Total Probability
The formula for the theorem of total probability is:
P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)
Substitute the values we found:
P(A) = \left( \frac{1}{2} \times \frac{7}{12} \right) + \left( \frac{1}{2} \times \frac{5}{12} \right)
P(A) = \frac{7}{24} + \frac{5}{24}
P(A) = \frac{12}{24}
P(A) = \frac{1}{2}
Final Answer: \boxed{ P(A) = \frac{1}{2} }
Q2. 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.
Solution:
Step 1: Define the events
Let E_1 be the event of selecting Bag I.
Let E_2 be the event of selecting Bag II.
Let A be the event that the ball drawn is Red.
We want to find the probability that the ball was drawn from the first bag, given that it is red, which is P(E_1|A).
Step 2: Find the probabilities of selecting each bag
P(E_1) = \frac{1}{2}
P(E_2) = \frac{1}{2}
Step 3: Find the conditional probabilities of drawing a red ball from each bag
Bag I contains 4 Red and 4 Black balls (Total 8).
The probability of drawing a Red ball given it is from Bag I is: P(A|E_1) = \frac{4}{8} = \frac{1}{2}
Bag II contains 2 Red and 6 Black balls (Total 8).
The probability of drawing a Red ball given it is from Bag II is: P(A|E_2) = \frac{2}{8} = \frac{1}{4}
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{\frac{1}{2} \times \frac{1}{2}}{\left( \frac{1}{2} \times \frac{1}{2} \right) + \left( \frac{1}{2} \times \frac{1}{4} \right)}
P(E_1|A) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}}
To simplify the denominator, find a common denominator (8):
P(E_1|A) = \frac{\frac{1}{4}}{\frac{2}{8} + \frac{1}{8}}
P(E_1|A) = \frac{\frac{1}{4}}{\frac{3}{8}}
P(E_1|A) = \frac{1}{4} \times \frac{8}{3} = \frac{8}{12}
P(E_1|A) = \frac{2}{3}
Final Answer: \boxed{ P(E_1|A) = \frac{2}{3} }
Q3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars 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 a hostlier?
Solution:
Step 1: Define the events
Let E_1 be the event that the student is a hostlier.
Let E_2 be the event that the student is a day scholar.
Let A be the event that the student attains an ‘A’ grade.
We want to find the probability that the student is a hostlier, given that he has an A grade, which is P(E_1|A).
Step 2: Given probabilities of being a hostlier or day scholar
Hostelier P(E_1) = 60 \% = 0.6
Day Scholar P(E_2) = 40 \% = 0.4
Step 3: Find the conditional probabilities of getting an ‘A’ grade
The probability of a hostlier attaining an ‘A’ grade is: P(A|E_1) = 30 \% = 0.3
The probability of a day scholar attaining an ‘A’ grade is: P(A|E_2) = 20 \% = 0.2
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{0.6 \times 0.3}{(0.6 \times 0.3) + (0.4 \times 0.2)}
P(E_1|A) = \frac{0.18}{0.18 + 0.08}
P(E_1|A) = \frac{0.18}{0.26}
P(E_1|A) = \frac{18}{26} = \frac{9}{13}
Final Answer: \boxed{ P(E_1|A) = \frac{9}{13} }
Q4. In answering a question on a multiple choice test, a student either 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 the student knows the answer given that he answered it correctly?
Solution:
Step 1: Define the events
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 answered correctly.
We want to find the probability that the student knows the answer, given that he answered correctly, which is P(E_1|A).
Step 2: Given probabilities
P(E_1) = \frac{3}{4}
P(E_2) = \frac{1}{4}
Step 3: Conditional probabilities of answering correctly
If the student knows the answer, he will answer correctly. The probability of answering correctly given he knows the answer is 1.
P(A|E_1) = 1
If the student guesses the answer, given the probability of answering correctly is:
P(A|E_2) = \frac{1}{4}
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{\frac{3}{4} \times 1}{\left( \frac{3}{4} \times 1 \right) + \left( \frac{1}{4} \times \frac{1}{4} \right)}
P(E_1|A) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}}
P(E_1|A) = \frac{\frac{3}{4}}{\frac{12}{16} + \frac{1}{16}}
P(E_1|A) = \frac{\frac{3}{4}}{\frac{13}{16}}
P(E_1|A) = \frac{3}{4} \times \frac{16}{13} = \frac{48}{52}
P(E_1|A) = \frac{12}{13}
Final Answer: \boxed{ P(E_1|A) = \frac{12}{13} }
Q5. 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?
Solution:
Step 1: Define the events
Let E_1 be the event that the person actually has the disease.
Let E_2 be the event that the person is healthy (does not have the disease).
Let A be the event that the test result is positive.
We want to find the probability that a person has a disease given that his result is positive P(E_1|A).
Step 2: Given Probabilities
0.1% of the population has the disease:
P(E_1) = 0.1 \% = 0.001
This means the rest of the population is healthy:
P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0.999
Step 3: Find the conditional probabilities of a positive test
The test is 99% effective if the disease is present:
P(A|E_1) = 99 \% = 0.99
The test gives a false positive for 0.5% of healthy people:
That is the test gives a positive result for health people (not having the disease)
P(A|E_2) = 0.5 \% = 0.005
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{0.001 \times 0.99}{(0.001 \times 0.99) + (0.999 \times 0.005)}
P(E_1|A) = \frac{0.00099}{0.00099 + 0.004995}
P(E_1|A) = \frac{0.00099}{0.005985}
Multiply numerator and denominator by 1,000,000 to remove decimals:
P(E_1|A) = \frac{990}{5985}
P(E_1|A) = \frac{198}{1197} = \frac{22}{133}
Final Answer: \boxed{ P(E_1|A) = \frac{22}{133} }
Q6. 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 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?
Solution:
Step 1: Define the events
Let E_1 be the event of choosing the two-headed coin.
Let E_2 be the event of choosing the biased coin.
Let E_3 be the event of choosing the unbiased coin.
Let A be the event that the toss of a coin shows heads.
We want to find P(E_1|A).
Step 2: Find the probabilities of selecting each coin
One of the three coins is chosen at random:
P(E_1) = \frac{1}{3}
P(E_2) = \frac{1}{3}
P(E_3) = \frac{1}{3}
Step 3: Find the conditional probabilities of getting heads
For the two-headed coin, heads is certain: P(A|E_1) = 1
For the biased coin, heads comes up 75% of the time: P(A|E_2) = 75 \% = \frac{3}{4}
For the unbiased coin, heads comes up 50% of the time: P(A|E_3) = \frac{1}{2}
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{\frac{1}{3} \times 1}{\left( \frac{1}{3} \times 1 \right) + \left( \frac{1}{3} \times \frac{3}{4} \right) + \left( \frac{1}{3} \times \frac{1}{2} \right)}
P(E_1|A) = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}}
P(E_1|A) = \frac{1}{\frac{4}{4} + \frac{3}{4} + \frac{2}{4}}
P(E_1|A) = \frac{1}{\frac{9}{4}}
P(E_1|A) = \frac{4}{9}
Final Answer: \boxed{ P(E_1|A) = \frac{4}{9} }
Q7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 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?
Solution:
Step 1: Define the events
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 person meets with an accident.
We have to find P(E_1|A).
Step 2: Find the probabilities of selecting each driver type
Total number of insured persons = 2000 + 4000 + 6000 = 12000
Scooter: P(E_1) = \frac{2000}{12000} = \frac{2}{12} = \frac{1}{6}
Car: P(E_2) = \frac{4000}{12000} = \frac{4}{12} = \frac{1}{3}
Truck: P(E_3) = \frac{6000}{12000} = \frac{6}{12} = \frac{1}{2}
Step 3: Given conditional probabilities of an accident:
P(A|E_1) = 0.01
P(A|E_2) = 0.03
P(A|E_3) = 0.15
Step 4: Apply 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)}
Substitute the values we found:
P(E_1|A) = \frac{\frac{1}{6} \times 0.01}{\left( \frac{1}{6} \times 0.01 \right) + \left( \frac{1}{3} \times 0.03 \right) + \left( \frac{1}{2} \times 0.15 \right)}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.03}{3} + \frac{0.15}{2}}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + 0.01 + 0.075}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + 0.01 + 0.075}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.01}{6} + \frac{0.06}{6} + \frac{0.45}{6}}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.01 + 0.06 + 0.45}{6}}
P(E_1|A) = \frac{\frac{0.01}{6}}{\frac{0.52}{6}}
P(E_1|A) = \frac{0.01}{0.52} = \frac{1}{52}
Final Answer: \boxed{ P(E_1|A) = \frac{1}{52} }
Q8. 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 from this and is found to be defective. What is the probability that it was produced by machine B?
Solution:
Step 1: Define the events
Let E_1 be the event that the item is produced by machine A.
Let E_2 be the event that the item is produced by machine B.
Let A be the event that the item chosen is defective.
We have to find the probability that the defective item was produced by machine B, i.e. P(E_2|A).
Step 2: Given probabilities of production by each machine
P(E_1) = 60 \% = 0.6
P(E_2) = 40 \% = 0.4
Step 3: Conditional probabilities of an item being defective
Given it’s from machine A is: P(A|E_1) = 2 \% = 0.02
Given it’s from machine B is: P(A|E_2) = 1 \% = 0.01
Step 4: Applying 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)}
Substituting the values:
P(E_2|A) = \frac{0.4 \times 0.01}{(0.6 \times 0.02) + (0.4 \times 0.01)}
P(E_2|A) = \frac{0.004}{0.012 + 0.004}
P(E_2|A) = \frac{0.004}{0.016}
P(E_2|A) = \frac{4}{16} = \frac{1}{4}
Final Answer: \boxed{ P(E_2|A) = \frac{1}{4} }
Q9. 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.
Solution:
Step 1: Define the events
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.
We have 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).
Step 2: Given probabilities
Probability that the first group wins P(E_1) = 0.6
Probability that the second group wins P(E_2) = 0.4
Step 3: Given conditional probabilities of introducing a new product
The probability of a new product given the first group wins is: P(A|E_1) = 0.7
The probability of a new product given the second group wins is: P(A|E_2) = 0.3
Step 4: Apply 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)}
Substituting the values:
P(E_2|A) = \frac{0.4 \times 0.3}{(0.6 \times 0.7) + (0.4 \times 0.3)}
P(E_2|A) = \frac{0.12}{0.42 + 0.12}
P(E_2|A) = \frac{0.12}{0.54}
P(E_2|A) = \frac{12}{54} = \frac{2}{9}
Final Answer: \boxed{ P(E_2|A) = \frac{2}{9} }
Q10. Suppose a girl throws a die. 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 1, 2, 3 or 4 with the die?
Solution:
Step 1: Define the events
Let E_1 be the event that the girl gets 1, 2, 3, or 4.
Let E_2 be the event that the girl gets 5 or 6.
Let A be the event that she obtained exactly one head.
We have to find the probability that she threw 1, 2, 3, or 4, given that she got exactly one head. This is P(E_1|A).
Step 2: Given Probabilities
P(E_1) = \frac{4}{6} = \frac{2}{3}
P(E_2) = \frac{2}{6} = \frac{1}{3}
Step 3: Finding conditional probabilities of getting exactly one head
If E_1 occurs (tosses a coin once): Sample space = {H, T}.
The probability of getting exactly one head is:P(A|E_1) = \frac{1}{2}
If E_2 occurs (tosses a coin three times): Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (8 outcomes).
The outcomes with exactly one head are {HTT, THT, TTH}. There are 3 such outcomes.
The probability of getting exactly one head is: P(A|E_2) = \frac{3}{8}
Step 4: Applying 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)}
Substituting the values:
P(E_1|A) = \frac{\frac{2}{3} \times \frac{1}{2}}{\left( \frac{2}{3} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{3}{8} \right)}
P(E_1|A) = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{8}}
P(E_1|A) = \frac{\frac{1}{3}}{\frac{8}{24} + \frac{3}{24}}
P(E_1|A) = \frac{\frac{1}{3}}{\frac{11}{24}}
P(E_1|A) = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}
Final Answer: \boxed{ P(E_1|A) = \frac{8}{11} }
Q11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Solution:
Step 1: Define the events
Let E_1 be the event that the item is produced by operator A.
Let E_2 be the event that the item is produced by operator B.
Let E_3 be the event that the item is produced by operator C.
Let A be the event that a defective item is produced.
We have to find the probability that the item was produced by A given that it is defective P(E_1|A).
Step 2: Given probabilities
P(E_1) = 50 \% = 0.5
P(E_2) = 30 \% = 0.3
P(E_3) = 20 \% = 0.2
Step 3: Given conditional probabilities of
Producing a defective item by operator A P(A|E_1) = 1 \% = 0.01
Producing a defective item by operator B P(A|E_2) = 5 \% = 0.05
Producing a defective item by operator C P(A|E_3) = 7 \% = 0.07
Step 4: Applying 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)}
Substituting the values:
P(E_1|A) = \frac{0.5 \times 0.01}{(0.5 \times 0.01) + (0.3 \times 0.05) + (0.2 \times 0.07)}
P(E_1|A) = \frac{0.005}{0.005 + 0.015 + 0.014}
P(E_1|A) = \frac{0.005}{0.034}
P(E_1|A) = \frac{5}{34}
Final Answer: \boxed{ P(E_1|A) = \frac{5}{34} }
Q12. 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.
Solution:
Step 1: Define the events
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 the two cards drawn from the remaining 51 cards are both diamonds.
We have to find of the lost card being a diamond given that both cards drawn were diamond i.e. P(E_1|A).
Step 2: Given probabilities
There are 13 diamonds and 39 non-diamonds in a 52-card pack.
Lost card is diamond P(E_1) = \frac{13}{52} = \frac{1}{4}
Lost card is not a diamond P(E_2) = \frac{39}{52} = \frac{3}{4}
Step 3: Finding the conditional probabilities of event A (Both cards drawn are diamonds)
If E_1 occurs (lost card was a diamond): The pack has 51 cards remaining: 12 diamonds and 39 non-diamonds.
The probability of drawing 2 diamonds (event A) is C(12, 2) / C(51, 2) .
i.e. P(A|E_1) = \frac{\frac{12 \times 11}{2 \times 1}}{\frac{51 \times 50}{2 \times 1}}
P(A|E_1) = \frac{12 \times 11}{51 \times 50}
P(A|E_1) = \frac{132}{2550}
If E_2 occurs (lost card was not a diamond): The pack has 51 cards remaining: 13 diamonds and 38 non-diamonds.
The probability of drawing 2 diamonds (event A) is C(13, 2) / C(51, 2) .
i.e. P(A|E_2) = \frac{\frac{13 \times 12}{2 \times 1}}{\frac{51 \times 50}{2 \times 1}}
P(A|E_2) = \frac{13 \times 12}{51 \times 50}
P(A|E_2) = \frac{156}{2550}
Step 4: Applying 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)}
Substituting the values:
P(E_1|A) = \frac{\frac{1}{4} \times \frac{132}{2550}}{\left( \frac{1}{4} \times \frac{132}{2550} \right) + \left( \frac{3}{4} \times \frac{156}{2550} \right)}
P(E_1|A) = \frac{132}{132 + (3 \times 156)}
P(E_1|A) = \frac{132}{132 + 468}
P(E_1|A) = \frac{132}{600}
P(E_1|A) = \frac{11}{50}
Final Answer: \boxed{ P(E_1|A) = \frac{11}{50} }
Q13. Probability that A speaks truth is \frac{4}{5}. A coin is tossed. A reports that a 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}
Solution:
Step 1: Define the events
Let E_1 be the event that a Head actually appears on the coin.
Let E_2 be the event that a Tail actually appears (Head does not).
Let A be the event that ‘A reports that a head appears’.
We have to find that head actually appears given that A reports it is a head i.e. P(E_1|A).
Step 2: Given Probabilities
Head appears P(E_1) = \frac{1}{2}
Tail appears P(E_2) = \frac{1}{2}
Step 3: Find the conditional probabilities of A’s report
If E_1 occurs (Head): A reporting “Head” means A is speaking the truth.
P(A|E_1) = P(\text{A speaks truth}) = \frac{4}{5}
If E_2 occurs (Tail): A reporting “Head” means A is lying (not speaking the truth).
P(A|E_2) = P(\text{A lies}) = 1 - \frac{4}{5} = \frac{1}{5}
Step 4: Applying 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)}
Substituting the values:
P(E_1|A) = \frac{\frac{1}{2} \times \frac{4}{5}}{\left( \frac{1}{2} \times \frac{4}{5} \right) + \left( \frac{1}{2} \times \frac{1}{5} \right)}
P(E_1|A) = \frac{\frac{4}{5}}{\frac{4}{5} + \frac{1}{5}}
P(E_1|A) = \frac{\frac{4}{5}}{\frac{5}{5}} = \frac{4/5}{1}
\boxed{ P(E_1|A) = \frac{4}{5}}
Correct option is (A).
Q14. If A and B are two events such that A \subset B and P(B) \neq 0, then which of the following is correct?
(A) P(A|B) = \frac{P(B)}{P(A)}
(B) P(A|B) < P(A)
(C) P(A|B) \ge P(A)
(D) None of these
Solution:
Given A \subset B (A is a subset of B).
This means that if event A occurs, event B must also occur.
Because A \subset B , the intersection of A and B is just A itself.
i.e. A \cap B = A.
\implies P(A \cap B) = P(A).
Evaluating P(A|B)
The formula for conditional probability is:
P(A|B) = \frac{P(A \cap B)}{P(B)}
Substitute P(A) for P(A \cap B):
P(A|B) = \frac{P(A)}{P(B)}[/latex]
Compare P(A|B) with P(A)
We have P(A|B) = \frac{P(A)}{P(B)}.
We also know that since A \subset B , it must be that P(A) \le P(B) .
Because probabilities are between 0 and 1, we know P(B) \le 1 .
Dividing by P(B) (which is non-zero) gives:
\frac{1}{P(B)} \ge 1
Now, let’s look at P(A|B) = P(A) \times \frac{1}{P(B)}.
Since \frac{1}{P(B)} \ge 1 (and P(A) \ge 0 ):
P(A) \times \frac{1}{P(B)} \ge P(A) \times 1
This means P(A|B) \ge P(A).
Let’s check the options:
(A) P(A|B) = \frac{P(B)}{P(A)} . This is incorrect. We found P(A|B) = \frac{P(A)}{P(B)} .
(B) P(A|B) < P(A) . This is incorrect. We found P(A|B) \ge P(A) .
(C) P(A|B) \ge P(A) . This matches our derivation.
The correct option is (C).