Bayes’ Theorem Questions

Bayes’ theorem questions with solutions are given here for students to practice and understand how to apply Bayes’ theorem as a special case for conditional probability. These questions are specifically designed as per the CBSE class 12 syllabus. Every year, a good weightage question is asked based on Bayes’ theorem; practicising these questions will help in the preparations for the board exams.

Famous mathematician Thomas Bayes gave this theorem to solve the problem of finding reverse probability by using conditional probability.

The theorem is stated as follows:

Bayes’ theorem for two events is given as:

To learn more about Bayes’ Theorem, click here.

Bayes’ Theorem Questions With Solutions

To get a better understanding of Bayes’ theorem, let us apply it by solving a few questions.

Question 1:

Three persons A, B and C have applied for a job in a private company. The chance of their selections is in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve the profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C.

Solution:

Let E₁: person A get selected

E₂: person B get selected

E₃: person C get selected

A: Changes introduced but profit not happened

Now, P(E₁) = 1/(1+2+4) = 1/7

P(E₂) = 2/7 and P(E₃) = 4/7

P(A|E₁) = P(Profit not happened by the changes introduces by A) = 1 – P(Profit happened by the changes introduces by A) = 1 – 0.8 = 0.2

P(A|E₂) = P(Profit not happened by the changes introduces by B) = 1 – P(Profit happened by the changes introduces by B) = 1 – 0.5 = 0.5

P(A|E₃) = P(Profit not happened by the changes introduces by C) = 1 – P(Profit happened by the changes introduces by C) = 1 – 0.3 = 0.7

We have to find the probability of not happening profit due to selection of C

= 7/10.

∴ the required probability is 0.7.

Question 2:

A bag contains 4 balls. Two balls are drawn at random without replacement and are found to be blue. What is the probability that all balls in the bag are blue?

Solution:

Let E₁ = Bag contains two blue balls

E₂ = Bag contains three blue balls

E₃ = Bag contains four blue balls

A = event of getting two white balls

P(E₁) = P(E₂) = P(E₃) = ⅓

P(A|E₁) = ²C₂/⁴C₂ = ⅙

P(A|E₂) = ³C₂/⁴C₂ = ½

P(A|E₃) = ⁴C₂/⁴C₂ = 1

= [⅓ × 1]/[⅓ × ⅙ + ⅓ × ½ + ⅓ × 1]

= ⅗.

Check your answers with Bayes’ theorem calculator.

Question 3:

In a neighbourhood, 90% children were falling sick due flu and 10% due to measles and no other disease. The probability of observing rashes for measles is 0.95 and for flu is 0.08. If a child develops rashes, find the child’s probability of having flu.

Solution:

Let,

F: children with flu

M: children with measles

R: children showing the symptom of rash

P(F) = 90% = 0.9

P(M) = 10% = 0.1

P(R|F) = 0.08

P(R|M) = 0.95

= 0.072/(0.095 + 0.072) = 0.072/0.167 ≈ 0.43

⇒ P(F|R) = 0.43

Question 4:

There are three identical cards except that both the sides of the first card is coloured red, both sides of the second card is coloured blue and for the third card one side is coloured red and the other side is blue. One card is randomly selected among these three cards and put down, visible side of the card is red. What is the probability that the other side is blue?

Solution:

Let RR: card with both side red

BB: card with both side blue

RB: card with one side red and other side blue

A: event that the visible side of the chosen card is red

P(RR) = ⅓ , P(BB) = ⅓ and P(RB) = ⅓

By the theorem of total probability, P(A) = P(A|RR).P(RR) + P(A|BB).P(BB) + P(A|RB).P(RB)

∴ P(A) = 1 × ⅓ + 0 × ⅓ + ½ × ⅓ = ½

Now,

= ½ × ⅓ ÷ ½ = ⅓ .

Question 5:

Three urns are there containing white and black balls; first urn has 3 white and 2 black balls, second urn has 2 white and 3 black balls and third urn has 4 white and 1 black balls. Without any biasing one urn is chosen from that one ball is chosen randomly which was white. What is probability that it came from the third urn?

Solution:

Let E₁ = event that the ball is chosen from first urn

E₂ = event that the ball is chosen from second urn

E₃ = event that the ball is chosen from third urn

A = event that the chosen ball is white

Then, P(E₁) = P(E₂) = P(E₃) = ⅓.

P(A|E₁) = 3/5

P(A|E₂) = ⅖

P(A|E₃) = ⅘

= 4/9.

Question 6:

It is observed that 50% of mails are spam. There is a software that filters spam mail before reaching the inbox. It accuracy for detecting a spam mail is 99% and chances of tagging a non-spam mail as spam mail is 5%. If a certain mail is tagged as spam find the probability that it is not a spam mail.

Solution:

Let E₁ = event of spam mail

E₂ = event of non-spam mail

A = event of detecting a spam mail

Now, P(E₁) = 0.5 and P(E₂) = 0.5

P(A|E₁) = 0.99 and P(A|E₂) = 0.05

Then,

= 5/104 ≈ 4.8%

Also Check:

• Multiplication Rule of Probability
• Probability for Class 12
• Probability Distribution
• Bionomial Distribution

Question 7:

An unbiased dice is rolled and for each number on the dice a bag is chosen:

Bag A contains 3 white ball and 2 black ball, bag B contains 3 white ball and 4 black ball and bag C contains 4 white ball and 5 black ball. Dice is rolled and bag is chosen, if a white ball is chosen find the probability that it is chosen from bag B.

Solution:

Let E₁ = event of choosing bag A

E₂ = event of choosing bag B

E₃ = event of choosing bag C

A = event of choosing white ball

Then, P(E₁) = ⅙, P(E₂) = 2/6 = ⅔, P(E₃) = 3/6 = ½

And P(A|E₁) = ⅗, P(A|E₂) = 3/7, P(A|E₃) = 4/9

⇒ P(E₂|A) = 90/293.

Question 8:

A insurance company has insured 4000 doctors, 8000 teachers and 12000 businessmen. The chances of a doctor, teacher and businessman dying before the age of 58 is 0.01, 0.03 and 0.05, respectively. If one of the insured people dies before 58, find the probability that he is a doctor.

Solution:

Let, E₁ = event of a person being a doctor

E₂ = event of a person being a teacher

E₃ = event of a person being a businessman

A = event of death of an insured person

P(E₁) = 4000/(4000+8000+12000) = ⅙

P(E₂) = 8000/(4000+8000+12000) = ⅓

P(E₃) = 12000/(4000+8000+12000) = ½

P(A|E₁) = 0.01, P(A|E₂) = 0.03 and P(A|E₃) = 0.05

Therefore,

⇒ P(E₁|A) = 1/22

Question 9:

A card is lost from a pack of 52 cards. From the remaining cards two are drawn randomly and found to be both clubs. Find the probability that the lost card is also a clubs.

Solution:

Let E₁ = Lost card is a club

E₂ = lost card is not a club

A = both drawn cards are clubs

P(E₁) = 13/52 = ¼ P(E₂) = 39/52 = ¾

P(A|E₁) = P(drawing both club cards when the lost card is a club) = 12/51 × 11/50

P(A|E₂) = P(drawing both club cards when the lost card is not a club) = 13/51 × 12/50

Now,

⇒ P(E₁|A) = 11/50.

Question 10:

In shop A, 30 tin pure ghee and 40 tin adulterated ghee are kept for sale while in shop B, 50 tin pure ghee and 60 tin adulterated ghee are there. One tin of ghee is purchased from one of the shops randomly and it is found to be adulterated. Find the probability that it is purchased from shop B.

Solution:

Let E₁ = event of choosing shop A

E₂ = event of choosing shop B

A = event of purchasing adultrated tin of ghee

P(E₁) = ½ and P(E₂) = ½

P(A|E₁) = P(purchasing adultrated ghee from shop A) = 40/70 = 4/7

P(A|E₂) = P(purchasing adultrated ghee from shop B) = 60/110 = 6/11

Therefore,

Practice Questions on Bayes’ Theorem

1. If A, B and C have chances of being selected as a manager at private firm is in the ratio 4:1:2. The chances of for them to introduce changes in marketing strategies are 0.3, 0.8 and 0.5, respectively. If a change has taken place, find the the probability that it is due to the selection of B.

2. A man speaks the truth 4 out of 5 times. He throws a die and reports that it is actually a six. Find the probability that it is actually a six.

3. A sack contains 4 balls. Two balls are drawn at random (without replacement) and are found to be red. What is the probability that all balls in the bag are red?

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