ML Aggarwal Class 12 Maths Solutions Section B Chapter 04 Probability

4.1 Bayes' Theorem

If E₁, E₂, E₃, …, Eₙ are mutually exclusive and exhaustive events associated with a random experiment and A is any event associated with the experiment, then

P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / ΣP(Eᵢ) P(A|Eᵢ), where i = 1, 2, 3, …, n.

Proof: By the law of total probability, we have

P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + … + P(Eₙ) P(A|Eₙ) = ΣP(Eᵢ) P(A|Eᵢ) ...(i)

Also by the multiplication law of probability, we have

P(A ∩ Eᵢ) = P(A) P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ), i = 1, 2, 3, …, n

⇒ P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / P(A), i = 1, 2, 3, …, n

⇒ P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / ΣP(Eᵢ) P(A|Eᵢ), i = 1, 2, 3, …, n (by using (i))

The probability P(Eᵢ|A) means finding the probability of event Eᵢ given that event A has occurred. Probability P(Eᵢ) was already known - so it was an a priori probability. P(Eᵢ|A) is to be calculated after the knowledge that event A has happened - so it is called posteriori probability.

For example, suppose that in a factory, 60% of products are made by machine M₁ and 40% by machine M₂. Machine M₁ makes 1% faulty items and machine M₂ makes 2% faulty items, and let

E₁ = event that product is made by machine M₁,
E₂ = event that product is made by machine M₂ and
A = event that product is faulty.

Then from given information, we have

P(E₁) = 0.60, P(E₂) = 0.40, P(A|E₁) = 0.01, P(A|E₂) = 0.02.

Using the law of total probability, we can work out that the probability of a product being faulty,

P(A) = P(E₁) · P(A|E₁) + P(E₂) · P(A|E₂) = (0.60)(0.01) + (0.40)(0.02) = 0.006 + 0.008 = 0.014.

Thus, 1.4% of the factory's products are faulty.

If the event A happens - that is, if we pick up a product and find that it is faulty - P(E₁|A) means finding the probability that it was made by machine M₁. You may think that it is 60% as we are given that 60% of products are made by machine M₁. However, according to Bayes' theorem,

P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(0.60)(0.01)] / [(0.60)(0.01) + (0.40)(0.02)] = 0.006 / (0.006 + 0.008) = 0.006 / 0.014 = 3/7, i.e. approximately 43%.

Thus if we pick up a product, there is a 60% chance that it came from machine M₁ and 40% chance that it came from machine M₂. However, when we examine the product and find it to be faulty - we revise our probabilities (from P(E₁) to P(E₁|A) etc.) and say that there is a 43% chance that this product came from machine M₁ and 57% chance that it came from machine M₂. Thus, an a priori probability of 60% is revised to posteriori probability of 43% with the extra information that the product is faulty. This is what real life is all about. When we receive extra information, we revise our first opinions.

Illustrative Examples

 

Example 1. Bag I contains 2 white and 3 red balls and bag II contains 4 white and 5 red balls. A bag is taken at random and a ball is drawn from it. If the ball drawn is red, find the probability that it was drawn from bag I.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = bag I is taken,
E₂ = bag II is taken and
A = ball drawn is red.
Then P(E₁) = 1/2 = P(E₂).
P(A|E₁) = P(drawing a red ball from bag I) = 3/5,
P(A|E₂) = P(drawing a red ball from bag II) = 5/9.
We want to find P(E₁|A).
By using Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(1/2)(3/5)] / [(1/2)(3/5) + (1/2)(5/9)] = (3/10) / [(3/10) + (5/18)] = (3/10) / [(27 + 25)/90] = (3/10) / (52/90) = (3/10) × (90/52) = 27/52.

Exam Tip: Always write out the events E₁, E₂ etc. clearly at the start, then identify the given probabilities and what you need to find. Use Bayes' formula directly, substituting values carefully.

 

Example 2. There are two groups of subjects, one of which consists of 5 science and 3 engineering subjects; and the other consists of 3 science and 5 engineering subjects. An unbiased die is rolled. If number 1 or 6 turn up, a subject is selected at random from the first group, otherwise a subject is selected from the second group. If ultimately an engineering subject is selected, find the probability that it is selected from the second group.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = die turns up with number 1 or 6 - that is, selecting first group of subjects,
E₂ = die turns up with number 2, 3, 4 or 5 - that is, selecting second group of subjects and
A = engineering subject is selected.
Then P(E₁) = 2/6 = 1/3 and P(E₂) = 4/6 = 2/3.
P(A|E₁) = P(selecting engineering subject from first group) = 3/8,
P(A|E₂) = P(selecting engineering subject from second group) = 5/8.
We want to find P(E₂|A).
By Bayes' theorem, we have
P(E₂|A) = [P(E₂) P(A|E₂)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(2/3)(5/8)] / [(1/3)(3/8) + (2/3)(5/8)] = (10/24) / [(3/24) + (10/24)] = (10/24) / (13/24) = 10/13.

Exam Tip: When a die is rolled to determine a choice between two groups, carefully identify which outcomes lead to which group. Calculate P(A|Eᵢ) for each group, then apply Bayes' formula.

 

Example 3. An insurance company insured 1500 scooter drivers, 2500 car drivers and 4500 truck drivers. The probability of a scooter, a car and a truck driver meeting with an accident is 0.01, 0.02 and 0.04 respectively. If one of the insured person meets with an accident, find the probability that he is a scooter driver.
Solution: Let E₁, E₂, E₃ and A be the events defined as follows:
E₁ = insured person is a scooter driver,
E₂ = insured person is a car driver,
E₃ = insured person is a truck driver and
A = person meets with an accident.
Total number of insured persons = 1500 + 2500 + 4500 = 8500.
Therefore, P(E₁) = 1500/8500 = 3/17, P(E₂) = 2500/8500 = 5/17, P(E₃) = 4500/8500 = 9/17 and
P(A|E₁) = P(a scooter driver meeting with an accident) = 0.01,
P(A|E₂) = 0.02, P(A|E₃) = 0.04.
We want to find P(E₁|A).
By Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃)] = [(3/17)(0.01)] / [(3/17)(0.01) + (5/17)(0.02) + (9/17)(0.04)] = [3 × 0.01] / [3 × 0.01 + 5 × 0.02 + 9 × 0.04] = 0.03 / 0.49 = 3/49.

Exam Tip: When there are three or more categories, calculate the prior probability of each category by dividing its count by the total. Then apply the full Bayes' formula with all terms in the denominator.

 

Example 4. There are two groups of bags. Group I consists of 3 bags, each containing 5 green balls and 3 blue balls. Group II consists of 2 bags, each containing 2 green and 4 blue balls. A bag has been selected and a ball has been drawn at random from one of the bags and is found to be green. Find the probability that this green ball has been drawn from a bag of group II.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = selecting a bag from group I,
E₂ = selecting a bag from group II and
A = green ball has been drawn.
Since there are 5 bags and group I consists of 3 bags and group II consists of 2 bags, therefore, P(E₁) = 3/5 and P(E₂) = 2/5.
If E₁ has occurred, then a bag from group I has been chosen. The bag chosen contains 5 green balls and 3 blue balls, therefore P(A|E₁) = 5/8.
Similarly P(A|E₂) = 2/6 = 1/3.
We want to find P(E₂|A).
By Bayes' theorem, we have
P(E₂|A) = [P(E₂) P(A|E₂)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(2/5)(1/3)] / [(3/5)(5/8) + (2/5)(1/3)] = (2/15) / [(15/40) + (2/15)] = (2/15) / [(3/8) + (2/15)] = (2/15) / [(45 + 16)/120] = (2/15) × (120/61) = 16/61.

Exam Tip: When each group contains multiple identical bags with the same composition, the probability of selecting from that group is the number of bags in the group divided by the total number of bags.

 

Example 5. A letter is known to have come either from TATANAGAR or KOLKATA. On the envelope, only the two consecutive letters TA are visible. What is the probability that the letter has come from (i) KOLKATA (ii) TATANAGAR.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = letter has come from KOLKATA,
E₂ = letter has come from TATANAGAR and
A = two consecutive visible letters are TA.
Letter can come either from KOLKATA or TATANAGAR, so P(E₁) = 1/2 = P(E₂).
The word KOLKATA has 7 letters, so there are 6 groups of two consecutive letters - KO, OL, LK, KA, AT, TA. Only one of these is 'TA'.
Therefore, P(A|E₁) = probability of event A when E₁ has occurred - that is, when letter has come from KOLKATA = 1/6.
The word TATANAGAR has 9 letters, so there are 8 groups of two consecutive letters - TA, AT, TA, AN, NA, AG, GA, AR. Two out of these are 'TA'.
Therefore, P(A|E₂) = probability of event A when E₂ has occurred - that is, when the letter has come from TATANAGAR = 2/8 = 1/4.
(i) We want to find P(E₁|A).
By Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(1/2)(1/6)] / [(1/2)(1/6) + (1/2)(1/4)] = (1/12) / [(1/12) + (1/8)] = (1/12) / [(2 + 3)/24] = (1/12) × (24/5) = 2/5.

Exam Tip: For questions about consecutive letters or arrangements, count all possible pairs carefully. The probability of seeing a specific pair depends on how many times it appears in the total number of consecutive positions.

 

(ii) We want to find P(E₂|A).
By Bayes' theorem, we have
P(E₂|A) = [P(E₂) P(A|E₂)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(1/2)(1/4)] / [(1/2)(1/6) + (1/2)(1/4)] = (1/8) / [(1/12) + (1/8)] = (1/8) / [(2 + 3)/24] = (1/8) × (24/5) = 3/5.

Exam Tip: Notice that P(E₁|A) + P(E₂|A) = 2/5 + 3/5 = 1, which is a good check on your work.

 

Example 6. For A, B and C the chances of being selected as the manager of a firm are 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8 and 0.5 respectively. If the change does takes place, find the probability that it is due to the appointment of B.
Solution: Let E₁, E₂, E₃ and E be the events as defined below:
E₁ = A is selected as manager,
E₂ = B is selected as manager,
E₃ = C is selected as manager and
E = radical change occurs in marketing strategy.
P(E₁) = 4/(4+1+2) = 4/7, P(E₂) = 1/(4+1+2) = 1/7 and P(E₃) = 2/(4+1+2) = 2/7.
Given P(E|E₁) = 0.3, P(E|E₂) = 0.8, P(E|E₃) = 0.5.
We want to find the probability that the radical change in marketing strategy has occurred due to the appointment of B - that is, we want to find P(E₂|E).
By Bayes' theorem, we have
P(E₂|E) = [P(E₂) P(E|E₂)] / [P(E₁) P(E|E₁) + P(E₂) P(E|E₂) + P(E₃) P(E|E₃)] = [(1/7)(0.8)] / [(4/7)(0.3) + (1/7)(0.8) + (2/7)(0.5)] = 0.8 / [1.2 + 0.8 + 1] = 0.8 / 3 = 8/30 = 4/15.

Exam Tip: When odds are given (like 4:1:2), convert them to probabilities by dividing each number by the sum. Then substitute into Bayes' formula carefully.

 

Example 7. A bag contains 3 green and 7 white balls. Two balls are drawn one by one at random without replacement. If the second ball drawn is green, what is the probability that the first ball drawn is also green?
Solution: Let E₁, E₂ and A be the events as defined below:
E₁ = first ball drawn is green,
E₂ = first ball drawn is white and
A = second ball drawn is green.
As the bag contains 3 green and 7 white balls, P(E₁) = 3/10 and P(E₂) = 7/10.
The second ball is drawn from the bag without replacement.
When E₁ has occurred - that is, when a green ball has been drawn - then the bag contains 9 balls out of which 2 are green and 7 are white, so
P(A|E₁) = probability of drawing second green ball when one green ball has already been drawn = 2/9.
When E₂ has occurred - that is, when a white ball has been drawn - then the bag contains 9 balls out of which 3 are green and 6 are white, so
P(A|E₂) = probability of drawing second green ball when one white ball has already been drawn = 3/9 = 1/3.
We want to find P(E₁|A).
By Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(3/10)(2/9)] / [(3/10)(2/9) + (7/10)(1/3)] = (6/90) / [(6/90) + (7/30)] = (2/30) / [(2/30) + (7/30)] = (2/30) / (9/30) = 2/9.

Exam Tip: In problems without replacement, after each draw the total number of balls and the number of each colour decreases by one. Be careful to update these counts for calculating conditional probabilities.

 

Example 8. Bag I contains 3 white and 4 black balls and bag II contains 4 white and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn at random from bag II. If the ball so drawn is white, find the probability that the transferred ball is black.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = black ball is transferred from bag I to bag II,
E₂ = white ball is transferred from bag I to bag II and
A = white ball has been drawn from bag II.
Then P(E₁) = 4/7 and P(E₂) = 3/7.
When E₁ has occurred - that is, when a black ball has been transferred from bag I to bag II - then bag II has 4 white and 6 black balls. Therefore, P(A|E₁) = probability of drawing a white ball from bag II when E₁ has occurred = 4/10 = 2/5.
When E₂ has occurred - that is, when a white ball has been transferred from bag I to bag II - then bag II has 5 white and 5 black balls. Therefore, P(A|E₂) = probability of drawing a white ball from bag II when E₂ has occurred = 5/10 = 1/2.
We want to find P(E₁|A).
By Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(4/7)(2/5)] / [(4/7)(2/5) + (3/7)(1/2)] = (8/35) / [(8/35) + (3/14)] = (8/35) / [(16 + 15)/70] = (8/35) × (70/31) = 16/31.

Exam Tip: When a ball is transferred from one bag to another, the receiving bag's composition changes by one ball. Calculate the conditional probabilities for the final draw based on the updated composition.

 

Example 9. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of heads. If she gets a 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 a 1, 2, 3 or 4 with the die?
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = girl gets 5 or 6 on throw of a die,
E₂ = girl gets 1, 2, 3 or 4 on throw of a die and
A = the girl gets exactly one head.
P(E₁) = 2/6 = 1/3 and P(E₂) = 4/6 = 2/3.
When E₁ has occurred, the girl tosses a coin three times and the sample space for this experiment is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Thus sample space contains 8 equally likely outcomes, out of which exactly one head occurs 3 times.
Therefore, P(A|E₁) = probability of getting exactly one head when E₁ has already occurred = 3/8.
When E₂ has occurred, the girl tosses a coin once and the sample space for this experiment = {H, T}.
Therefore, P(A|E₂) = probability of getting exactly one head when E₂ has already occurred = 1/2.
We want to find P(E₂|A).
By Bayes' theorem, we get
P(E₂|A) = [P(E₂) P(A|E₂)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(2/3)(1/2)] / [(1/3)(3/8) + (2/3)(1/2)] = (1/3) / [(1/8) + (1/3)] = (1/3) / [(3 + 8)/24] = (1/3) × (24/11) = 8/11.

Exam Tip: When the experiment changes based on the first event (fewer coin tosses vs more coin tosses), carefully identify the sample space and count outcomes for each scenario to find P(A|Eᵢ).

 

Example 10. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = die shows six - that is, six has occurred,
E₂ = die does not show six - that is, six has not occurred and
A = the man reports that six has occurred.
We wish to calculate the probability that six has actually occurred given that the man reports that six occurs - that is, P(E₁|A).
Now, P(E₁) = 1/6, P(E₂) = 5/6,
P(A|E₁) = probability that the man reports that six occurs given that six has occurred = probability that the man is telling the truth = 3/4 and
P(A|E₂) = probability that the man reports that six occurs given that six has not occurred = probability that the man does not speak truth = 1/4.
By Bayes' theorem,
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(1/6)(3/4)] / [(1/6)(3/4) + (5/6)(1/4)] = (3/24) / [(3/24) + (5/24)] = (3/24) / (8/24) = 3/8.

Exam Tip: When dealing with truth-telling or accuracy problems, P(A|E₁) is the accuracy when the event is true (truth-telling probability), and P(A|E₂) is the probability of giving that report when the event is false (1 - accuracy).

 

Example 11. In a test, an examinee either guess or copies or knows the answer to a multiple choice question with four choices and only one correct option. The probability that he makes a guess is 1/3. The probability that he copies the answer is 1/6. The probability that the answer is correct, given that he copied it, is 1/8. Find the probability that he knows the answer to the question, given that he correctly answered it.
Solution: Let E₁, E₂, E₃ and A be the events defined as follows:
E₁ = the examinee guesses the answer,
E₂ = the examinee copies the answer,
E₃ = the examinee knows the answer and
A = the examinee has answered the question correctly.
P(E₁) = 1/3, P(E₂) = 1/6 (given).
As E₁, E₂ and E₃ are mutually exclusive and exhaustive events, P(E₁) + P(E₂) + P(E₃) = 1
⇒ P(E₃) = 1 - P(E₁) - P(E₂) = 1 - 1/3 - 1/6 = 1/2.
When E₁ has occurred, then the examinee guesses. Since there are four choices and only one is correct, the probability that he answers correctly given that he has made a guess is 1/4 - that is, P(A|E₁) = 1/4.
P(A|E₂) = probability that he answers correctly given that he has copied = 1/8.
When E₃ has occurred - that is, the examinee knows the answer - then the probability that he answers correctly given that he knows the answer is 1 (sure event) - that is, P(A|E₃) = 1.
We want to find P(E₃|A).
By Bayes' theorem, we have
P(E₃|A) = [P(E₃) P(A|E₃)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃)] = [(1/2)(1)] / [(1/3)(1/4) + (1/6)(1/8) + (1/2)(1)] = (1/2) / [(1/12) + (1/48) + (1/2)] = (1/2) / [(4 + 1 + 24)/48] = (1/2) × (48/29) = 24/29.

Exam Tip: For MCQ problems where an examinee guesses, knows, or copies: when guessing with 4 options, P(correct|guess) = 1/4; when copying from a questionable source, use the given probability; when knowing, P(correct|knows) = 1.

 

Example 12. A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white?
Solution: Let E₁, E₂, E₃ and A be the events defined as follows:
E₁ = the bag contains 2 white balls and 2 non-white balls
E₂ = the bag contains 3 white balls and one non-white ball
E₃ = the bag contains all four white balls and
A = two white balls have been drawn from the bag.
Then P(E₁) = 1/3, P(E₂) = 1/3, P(E₃) = 1/3
P(A|E₁) = probability of drawing 2 white balls when E₁ has occurred - that is, the bag contains 2 white and 2 non-white balls = C(2,2) / C(4,2) = 1/6.
P(A|E₂) = probability of drawing 2 white balls when E₂ has occurred - that is, the bag contains 3 white balls and one non-white ball = C(3,2) / C(4,2) = 3/6 = 1/2 and
P(A|E₃) = probability of drawing 2 white balls when E₃ has occurred - that is, the bag contains all four white balls = C(4,2) / C(4,2) = 1.
We want to find P(E₃|A)
By Bayes' theorem, we get
P(E₃|A) = [P(E₃) P(A|E₃)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃)] = [(1/3)(1)] / [(1/3)(1/6) + (1/3)(1/2) + (1/3)(1)] = (1/3) / [(1/18) + (1/6) + (1/3)] = (1/3) / [(1 + 3 + 6)/18] = (1/3) / (10/18) = (1/3) × (18/10) = 6/10 = 3/5.
Hence, the required probability = 3/5.

Exam Tip: When the bag's composition is not fully known but there are a few possibilities, assign equal prior probabilities to each possibility (unless stated otherwise). Then calculate the conditional probability for the observed outcome using combinations.

 

Example 13. Coloured balls are distributed in three bags as shown in the following table:

A bag is selected at random and then two balls are drawn from the selected bag. They happen to be black and red. What is the probability that they have come from bag III?
Solution: Let E₁, E₂, E₃ and A be the events defined as follows:
E₁ = bag I is selected,
E₂ = bag II is selected,
E₃ = bag III is selected and
A = one black and one red ball has been drawn from the selected bag.
As the bags are selected at random, P(E₁) = P(E₂) = P(E₃) = 1/3.
Two balls are drawn randomly from the selected bag.
Total number of balls in bag I = 1 + 2 + 3 = 6.
P(A|E₁) = probability of drawing one black and one red ball when bag I has been selected = [C(1,1) × C(3,1)] / C(6,2) = (1 × 3) / 15 = 1/5.
Total number of balls in bag II = 2 + 4 + 1 = 7.
P(A|E₂) = probability of drawing one black and one red ball when bag II has been selected = [C(2,1) × C(1,1)] / C(7,2) = (2 × 1) / 21 = 2/21.
Total number of balls in bag III = 4 + 5 + 3 = 12.
P(A|E₃) = probability of drawing one black and one red ball when bag III has been selected = [C(4,1) × C(3,1)] / C(12,2) = (4 × 3) / 66 = 2/11.
We want to find P(E₃|A).
By Bayes' theorem, we have
P(E₃|A) = [P(E₃) P(A|E₃)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃)] = [(1/3)(2/11)] / [(1/3)(1/5) + (1/3)(2/21) + (1/3)(2/11)] = (2/33) / [(1/15) + (2/63) + (2/33)].
Finding a common denominator for the fractions 1/15, 2/63, 2/33:
LCM(15, 63, 33) = 1155
1/15 = 77/1155, 2/63 = (2 × 110)/1155 = 220/1155, wait let me recalculate this more carefully.
1/15 = 77/1155, 2/63 = (2 × 1155)/(63 × 1155) ... actually let's simplify.
(2/33) / [(1/15) + (2/63) + (2/33)] = (2/11) / [(1/5) + (2/21) + (2/11)].
To add 1/5 + 2/21 + 2/11, find LCM(5, 21, 11) = 1155
1/5 = 231/1155, 2/21 = 110/1155, 2/11 = 210/1155
Sum = (231 + 110 + 210)/1155 = 551/1155
So P(E₃|A) = (2/11) / (551/1155) = (2/11) × (1155/551) = (2 × 105)/551 = 210/551.

Exam Tip: When finding one of one colour and one of another, use the product of two separate binomial coefficients in the numerator. Always simplify fractions before substituting into Bayes' formula to avoid arithmetic errors.

 

Example 14. 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 clubs. Find the probability of the lost card being a club.
Solution: Let E₁, E₂ and A be the events defined as follows:
E₁ = lost card is of clubs,
E₂ = lost card is not of clubs and
A = two cards drawn are both of clubs.
Then P(E₁) = 13/52 = 1/4 and P(E₂) = 39/52 = 3/4.
When one card is lost, number of remaining cards in the pack = 51.
When E₁ has occurred - that is, a card of clubs is lost - then the probability of drawing 2 cards of clubs from the remaining pack = C(12,2) / C(51,2),
so P(A|E₁) = C(12,2) / C(51,2) = [(12 × 11)/(1 × 2)] / [(51 × 50)/(1 × 2)] = 132/2550 = 22/425.
When E₂ has occurred - that is, when a card of clubs is not lost - then the probability of drawing 2 cards of clubs from the remaining pack = C(13,2) / C(51,2),
so P(A|E₂) = C(13,2) / C(51,2) = [(13 × 12)/(1 × 2)] / [(51 × 50)/(1 × 2)] = 156/2550 = 26/425.
We want to find P(E₁|A).
By Bayes' theorem, we have
P(E₁|A) = [P(E₁) P(A|E₁)] / [P(E₁) P(A|E₁) + P(E₂) P(A|E₂)] = [(1/4)(22/425)] / [(1/4)(22/425) + (3/4)(26/425)] = (22/1700) / [(22/1700) + (78/1700)] = (22/1700) / (100/1700) = 22/100 = 11/50.

Exam Tip: When a card is removed from a deck, both the total number of remaining cards (51) and the number of the specific suit remaining change. Use these updated values in the combination formula.

 

Exercise 4.1

 

Question 1. Bag A contains 2 white and 4 red balls and bag B contains 5 white and 3 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.

Exam Tip: Define events clearly as selecting each bag and drawing red. Calculate P(red) using the law of total probability first if needed, then apply Bayes' theorem.

 

Question 2. A bag contains 4 red and 4 black balls. Another bag contains 2 red and 6 black balls. One of the bags is selected at random and a ball is drawn from the bag. If the ball is red, what is the probability that it came from the first bag?

Exam Tip: Both bags are equally likely to be selected. For each bag, identify the probability of drawing red. Apply Bayes' formula directly.

 

Question 3. A box contains 3 blue and 2 red balls while another box contains 2 blue and 5 red balls. A box is picked up randomly and a ball is drawn out.
(i) What are the chances of getting a blue ball?
(ii) If the ball drawn is blue, what is the probability that it came from the first box?

Exam Tip: Part (i) uses the law of total probability. Part (ii) uses Bayes' theorem. The probability from part (i) may be useful as a check.

 

Question 4. 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 selected student is a girl?

Exam Tip: Count the total number of rank holders (7). The probability of being a girl given rank holder status uses Bayes' theorem with the number of girls who are rank holders as the favorable numerator.

 

Question 5. Suppose 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 selected. Assume that there are equal number of men and women.

Exam Tip: Convert the given fractions to probabilities of being a good orator given each gender. The prior probability of each gender is 1/2. Apply Bayes' theorem.

 

Question 6. There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the rooms laughed loudly. What is the probability that the girl who laughed loudly was from room number 2?

Exam Tip: Count the total girls in each room and the total girls overall. The event "a girl laughed" means a girl from one of the rooms. Use Bayes' theorem with this as the given condition.

 

Question 7. It is known that 60% students in a college reside in hostel while remaining 40% are day scholars. 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade. If one randomly chosen student has A grade, what is the probability that he lives in the hostel?

Exam Tip: Convert percentages to decimal probabilities. P(hostel) = 0.6, P(A grade | hostel) = 0.3, etc. Apply Bayes' theorem for the posterior probability.

 

Question 8. A company has two plants to manufacture scooters. Plant I manufactures 80% of the scooters and plant II manufactures 20%. At plant I, 85 out of 100 scooters are rated as of standard quality and at plant II, 65 out of 100 scooters are rated as of standard quality. A scooter is chosen at random and is found to be of standard quality. What is the probability that it has come from plant I?

Exam Tip: Convert ratios to probabilities: P(standard | Plant I) = 0.85, P(standard | Plant II) = 0.65. Apply Bayes' formula with the given production percentages.

 

Question 9. In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In this class, 60% of the students are boys. If a student is selected at random and found to have an IQ of more than 150, find the probability that the student is a boy.

Exam Tip: P(boy) = 0.6, P(girl) = 0.4. P(IQ > 150 | boy) = 0.05, P(IQ > 150 | girl) = 0.10. Apply Bayes' theorem.

 

Question 10. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer. Assuming that the student who guesses at the answer will be correct with a probability 1/4, what is the probability that he knew the answer given that he answered it correctly?

Exam Tip: When the student knows the answer, P(correct | knows) = 1. When guessing, P(correct | guesses) = 1/4. Apply Bayes' theorem.

 

Question 11(i). Anshul speaks truth 4 times out of 5. A coin is tossed. Anshul reports that it is a head. What is the probability that it is actually a head?

Exam Tip: P(H) = 1/2, P(truth) = 4/5, so P(reports H | H) = 4/5 and P(reports H | T) = 1/5. Apply Bayes' theorem.

 

Question 11(ii). A speaks truth 8 times out of 10 times. A die is rolled. He reports that it was 5. What is the probability that it was actually 5?

Exam Tip: P(5 on die) = 1/6, P(reports 5 | 5 occurred) = 8/10, P(reports 5 | 5 did not occur) = 2/10 × (1/5) since there are 5 other outcomes. Apply Bayes' theorem.

 

Question 11(iii). A man is known to speak truth 3 times out of 5 times. He throws a die and reports that it a number greater than 4. Find the probability that it is actually a number greater than 4.

Exam Tip: Numbers greater than 4 are 5 and 6, so P(>4) = 2/6 = 1/3. P(reports >4 | >4) = 3/5 and P(reports >4 | not >4) = 2/5 × (1/4) since there are 4 other outcomes.

 

4.2 Random Variables / Functions

A random variable is often described as a variable whose values are determined by chance - that is, by the outcomes of a random experiment.

If to each point of the sample space we assign a real number, we then have a function defined on the sample space. It is called a random function or stochastic function. Loosely, it is called a random variable or stochastic variable. Thus, a random variable is a function whose domain is the sample space of a random experiment and whose range is a subset of real numbers.

For example, let S be the sample space of a simultaneous throw of two coins. Then S = {HH, HT, TH, TT}.

Let X denote the number of heads. Then X is a random variable. It can have values 0, 1 and 2 only. Thus, the domain of X is S and the range is {0, 1, 2}. Note that X(HT) = 1, X(TT) = 0 etc.

Let Y denote the number of tails. Then Y is also a random variable, having domain S and range {0, 1, 2}.

Note that though domain and range of X and Y is the same, X ≠ Y because X(HH) = 2 while Y(HH) = 0. Also note that many other random variables can be defined over this sample space; for example, number of heads minus number of tails, square of the number of heads etc.

Some other examples of random variables are:

(i) Three balls are drawn simultaneously from a bag containing 5 red, 3 black and 4 white balls. Let X be a variable defined as the number of white balls drawn. Then X is a random variable with domain = sample space S = {W₃, R₃, B₃, W₂R₁, W₂B₁, R₂W₁, R₂B₁, B₂R₁, B₂W₁, B₁R₁W₁} where W₁, W₂, W₃ denote one, two, three white balls respectively etc. and range = {0, 1, 2, 3}.

Note that X(W₃) = 3, X(W₂B₁) = 2, X(R₂B₁) = 0, X(R₃) = 0 etc. The random function X is shown in the adjoining diagram.

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