Bayes' Theorem Practice Questions

Bayes' Theorem is a mathematical formula used to calculate the probability of an event based on new information or evidence. It updates the probability of an event by combining prior knowledge with the new evidence, making it useful for decision-making under uncertainty.

Question 1: A person has undertaken a job. The probability of completing the job on time if it rains is 0.44, and the probability of completing the job on time if it does not rain is 0.95. If the probability that it will rain is 0.45, then determine the probability that the job will be completed on time.

Let:

• R: event that it rains
• R^c: event that it does not rain
• C: event that the job is completed on time

We are given:

P(R)=0.45,P(Rc)=1−0.45=0.55

P(C∣R)=0.44,P(C∣Rc)=0.95

By the law of total probability:

P(C)=P(R)P(C∣R)+P(Rc)P(C∣Rc)

Substitute values:

P(C)=(0.45)(0.44)+(0.55)(0.95)

P(C)=0.198+0.5225=0.7205

Question 2: There are three urns containing 3 white and 2 black balls, 2 white and 3 black balls, and 1 black and 4 white balls, respectively. There is an equal probability of each urn being chosen. One ball is equal probability chosen at random. What is the probability that a white ball will be drawn?

Solution:

Let E₁, E₂, and E₃ be the events of choosing the first, second, and third urn respectively. Then,
P(E₁) = P(E₂) = P(E₃) = 1/3

Let E be the event that a white ball is drawn. Then,
P(E/E₁) = 3/5, P(E/E₂) = 2/5, P(E/E₃) = 4/5

By theorem of total probability, we have
P(E) = P(E/E₁) . P(E₁) + P(E/E₂) . P(E₂) + P(E/E₃) . P(E₃)
⇒ P(E) = (3/5 × 1/3) + (2/5 × 1/3) + (4/5 × 1/3)
⇒ P(E) = 9/15 = 3/5

Question 3: 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 hearts. Find the probability of the lost card being a heart.

Solution:

Let E₁, E₂, E_3, and E₄ be the events of losing a card of hearts, clubs, spades, and diamonds respectively.
Then P(E₁) = P(E₂) = P(E₃) = P(E₄) = 13/52 = 1/4.

Let E be the event of drawing 2 hearts from the remaining 51 cards. Then,

P(E|E₁) = probability of drawing 2 hearts, given that a card of hearts is missing
⇒ P(E|E₁) = ¹²C₂ / ⁵¹C₂ = (12 × 11)/2! × 2!/(51 × 50) = 22/425

P(E|E₂) = probability of drawing 2 clubs ,given that a card of clubs is missing
⇒ P(E|E₂) = ¹³C₂ / ⁵¹C₂ = (13 × 12)/2! × 2!/(51 × 50) = 26/425

P(E|E₃) = probability of drawing 2 spades ,given that a card of hearts is missing
⇒ P(E|E₃) = ¹³C₂ / ⁵¹C₂ = 26/425

P(E|E₄) = probability of drawing 2 diamonds ,given that a card of diamonds is missing
⇒ P(E|E₄) = ¹³C₂ / ⁵¹C₂ = 26/425

Therefore,

P(E₁|E) = probability of the lost card is being a heart, given the 2 hearts are drawn from the remaining 51 cards

⇒ P(E₁|E) = P(E₁) . P(E|E₁)/P(E₁) . P(E|E₁) + P(E₂) . P(E|E₂) + P(E₃) . P(E|E₃) + P(E₄) . P(E|E₄)
⇒ P(E₁|E) = (1/4 × 22/425) / {(1/4 × 22/425) + (1/4 × 26/425) + (1/4 × 26/425) + (1/4 × 26/425)}
⇒ P(E₁|E) = 22/100 = 0.22

Hence, The required probability is 0.22.

Question 4: Suppose 15 men out of 300 men and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is selected.

Solution:

Given,

• Total Men = 300
• Total Women = 1000
• Good Orators among Men = 15
• Good Orators among Women = 25

Total number of good orators = 15 (from men) + 25 (from women) = 40
Probability of selecting a male orator:
P(Male Orator) = Numbers of male orators / total no of orators = 15/40 = 3/8

Question 5: A man is known to speak the lies 1 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: 

In a throw of a die, let
E₁ = event of getting a six,
E₂ = event of not getting a six and
E = event that the man reports that it is a six.

Then, P(E₁) = 1/6, and P(E₂) = (1 - 1/6) = 5/6

P(E|E₁) = probability that the man reports that six occurs when six has actually occurred 
⇒ P(E|E₁) = probability that the man speaks the truth
⇒ P(E|E₁) = 3/4

P(E|E₂) = probability that the man reports that six occurs when six has not actually occurred

⇒ P(E|E₂) = probability that the man does not speak the truth
⇒ P(E|E₂) = (1 - 3/4) = 1/4

Probability of getting a six, given that the man reports it to be six 

P(E₁|E) = P(E|E₁) × P(E₁)/P(E|E₁) × P(E₁) + P(E|E₂) × P(E₂)     [by Bayes theorem]
⇒ P(E₁|E) = (3/4 × 1/6)/{(3/4 × 1/6) + (1/4 × 5/6)}
⇒ P(E₁|E) = (1/8 × 3) = 3/8

Hence the probability required is 3/8.

Practice Problem

Question 1: A medical test for a disease is 95% accurate in detecting the disease (True Positive Rate). The probability of a person having the disease is 0.01 (1%). If a person tests positive for the disease, what is the probability that they actually have the disease? (Assume that the false positive rate is 5%).

Question 2: A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random, and one of them is red. What is the probability that the second ball drawn is also red, given that the first ball was red?

Question 3: In a factory, 80% of the products are produced by Machine A and 20% by Machine B. Machine A produces 2% defective items, while Machine B produces 5% defective items. If a product is found to be defective, what is the probability that it was produced by Machine A?

Question 4: A survey shows that 70% of people like ice cream, and 40% of people like both ice cream and chocolate. What is the probability that a person likes chocolate, given that they like ice cream?

Answer:-

1. 16.1%.
2. 33.33%.
3. 61.5%.
4. 57.1%.