Mastering Bayes’ theorem is crucial for students studying CBSE class 12 syllabus, as it carries significant weightage in the board exams. To help you grasp this concept, we've prepared a set of practice questions on Bayes’ theorem, which is a special case of conditional probability . Bayes’ theorem, named after the renowned mathematician Thomas Bayes, addresses the problem of determining reverse probability using conditional probability.

Here's how the theorem is stated:

Assume that E1, E2, E3, …, En are non-empty events that form a partition of the sample space S. This means that these events are pairwise disjoint and the union of E1, E2, E3, …, En equals S. If A is an event with non-zero probability that occurs with some Ei (where i = 1, 2, 3, …,n), then

Maths Notes Free PDFs

\(\begin{array}{l}P(E_{i}|A)=\frac{P(A|E_{i}).P(E_{i})}{\sum_{i=1}^{n}P(A|E_{i}).P(E_{i})};\:\:i=1,2,3,…,n.\end{array} \)

The Bayes’ theorem for two events is illustrated as follows:

To delve deeper into Bayes’ Theorem, click here .

Bayes' Theorem Applications 

Bayes' Theorem is a powerful tool in probability and is used in many real-world situations. It helps us update our beliefs when we get new information. The idea of Bayesian inference, which comes from Bayes' Theorem, is used in many areas like medicine, science, law, and even sports.

Here are some simple examples of how Bayes’ Theorem is used:

• Medical Testing:
  It helps doctors find the actual chance of a person having a disease if they test positive. Even if a test says “positive,” Bayes’ Theorem helps check how reliable that result is.

• Spam Email Filters:
  Email apps use Bayes' Theorem to check whether a message is spam by looking at the words used in the email. If certain spam-like words appear, the email is more likely to be marked as spam.

• Weather Forecasting:
  It helps weather apps update the chance of rain when new data (like wind or humidity) is received. So, the forecast becomes more accurate as more data comes in.

Artificial Intelligence & Machine Learning:
One important use is in Naïve Bayes classifiers, which help computers make decisions or predictions, like suggesting movies or predicting if a message is spam.

Let's solve a few questions to better understand Bayes’ theorem.

Question 1: Selection and Success Probability

Three people – Ramesh, Suresh, and Mahesh – applied for a position at a company. The chances of them getting selected are in the ratio 2:3:5. If selected, their chances of improving the company's performance are 0.7, 0.6, and 0.4 respectively. If no improvement is observed, what is the probability that Mahesh was selected?

Solution:

Let:
E₁ = Ramesh is selected
E₂ = Suresh is selected
E₃ = Mahesh is selected
A = No improvement in performance

• P(E₁) = 2 / (2+3+5) = 2/10

• P(E₂) = 3/10

• P(E₃) = 5/10

Now,
P(A|E₁) = 1 - 0.7 = 0.3
P(A|E₂) = 1 - 0.6 = 0.4
P(A|E₃) = 1 - 0.4 = 0.6

Using Bayes' Theorem:
P(E₃|A) = [P(E₃) × P(A|E₃)] / [P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + P(E₃)P(A|E₃)]

= (5/10 × 0.6) / [(2/10 × 0.3) + (3/10 × 0.4) + (5/10 × 0.6)]
= 0.3 / [0.06 + 0.12 + 0.3] = 0.3 / 0.48 = 0.625

Answer: 0.625 or 62.5% chance it was Mahesh.

Question 2: Ball Colors in a Bag

There’s a bag which might contain 2, 3, or 4 green balls. One bag is randomly chosen. Two balls are taken out without replacement, and both are green. What is the probability that all four balls are green?

Solution:

Let:
E₁ = Bag has 2 green balls
E₂ = Bag has 3 green balls
E₃ = Bag has 4 green balls
A = Both drawn balls are green

P(E₁) = P(E₂) = P(E₃) = 1/3

P(A|E₁) = 2C2 / 4C2 = 1 / 6
P(A|E₂) = 3C2 / 4C2 = 3 / 6 = 0.5
P(A|E₃) = 4C2 / 4C2 = 1

P(E₃|A) = (1/3 × 1) / [(1/3 × 1/6) + (1/3 × 0.5) + (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) = 0.6

Answer: 0.6 or 60% chance all balls are green.

Question 3: Flu vs. Cold with Rashes

Out of all students in a school, 85% are sick due to cold and 15% due to chickenpox. Rash is a symptom that appears in 0.1 of cold cases and 0.95 of chickenpox cases. If a student has a rash, what is the probability they have cold?

Solution:

Let:
C = Cold
P = Chickenpox
R = Rash

P(C) = 0.85, P(P) = 0.15
P(R|C) = 0.1
P(R|P) = 0.95

Using Bayes’ Theorem:

P(C|R) = [P(C) × P(R|C)] / [P(C) × P(R|C) + P(P) × P(R|P)]
= (0.85 × 0.1) / [(0.85 × 0.1) + (0.15 × 0.95)]
= 0.085 / (0.085 + 0.1425) = 0.085 / 0.2275 ≈ 0.3736

Answer: ≈ 37.36% chance the rash is due to cold.

Question 4: Colored Card Faces

Three cards:

• One is red on both sides (RR)

• One is blue on both sides (BB)

• One has red and blue (RB)

One card is picked and placed face-up. The visible side is red. What is the probability that the other side is blue?

Solution:

Possible red-face situations:

• RR: 2 red sides → 2 red sides out of 6

• RB: 1 red side → 1 red side out of 6

• BB: 0 red sides

So total red faces = 3 (2 from RR, 1 from RB)

P(Red side from RR) = 2/6
P(Red side from RB) = 1/6
Total P(Red) = 3/6 = 0.5

Now, probability that it’s RB given red side is visible:

P(RB|Red) = (1/6) / (3/6) = 1/3

Answer: 1/3 chance the other side is blue.

Question 5: White Ball from Urns

There are 3 urns:

• Urn 1: 2 white, 3 black

• Urn 2: 4 white, 1 black

• Urn 3: 3 white, 2 black

One urn is selected randomly. A white ball is drawn. What is the probability it came from Urn 2?

Solution:

Let:
E₁ = Choose Urn 1
E₂ = Choose Urn 2
E₃ = Choose Urn 3
A = Draw white ball

P(E₁) = P(E₂) = P(E₃) = 1/3
P(A|E₁) = 2/5
P(A|E₂) = 4/5
P(A|E₃) = 3/5

Now,

P(E₂|A) = [P(E₂) × P(A|E₂)] / [P(E₁) × P(A|E₁) + P(E₂) × P(A|E₂) + P(E₃) × P(A|E₃)]

= (1/3 × 4/5) / [(1/3 × 2/5) + (1/3 × 4/5) + (1/3 × 3/5)]
= (4/15) / (2/15 + 4/15 + 3/15) = 4/15 ÷ 9/15 = 4/9

Answer: 4/9 probability the white ball came from Urn 2.

Question 6:

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

Solution:

Let's denote the following events:

E1 = The person is a doctor

E2 = The person is a teacher

E3 = The person is a businessman

A = The death of an insured person

We can calculate the following probabilities:

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, P(A|E3) = 0.05

Therefore,

\(\begin{array}{l}P(E_{1}|A)=\frac{P(A|E_{1})P(E_{1})}{P(A|E_{1})P(E_{1})+P(A|E_{2})P(E_{2})+P(A|E_{3})P(E_{3})}\end{array} \)

\(\begin{array}{l}=\frac{0.01\times 1/6}{0.01\times 1/6+0.03\times 1/3+0.05\times 1/2}=\frac{1}{22}\end{array} \)

Thus, P(E1|A) = 1/22.

Question 7:

A card is lost from a deck of 52 cards. Two cards are drawn at random from the remaining cards and found to be both clubs. What is the probability that the lost card is also a club?

Solution:

Let's denote the following events:

E1 = The lost card is a club

E2 = The lost card is not a club

A = Both drawn cards are clubs

We can calculate the following probabilities:

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, we can calculate:

\(\begin{array}{l}P(E_{1}|A)=\frac{P(A|E_{1})P(E_{1})}{P(A|E_{1})P(E_{1})+P(A|E_{2})P(E_{2})}\end{array} \)

\(\begin{array}{l}=\frac{12/51 \times 11/50 \times 1/4}{12/51 \times 11/50 \times 1/4+13/51 \times 12/50 \times 3/4}=\frac{12\times 11}{12\times11+3\times 13\times 12}\end{array} \)

Therefore, P(E1|A) = 11/50.

Question 8:

Shop A has 30 tins of pure ghee and 40 tins of adulterated ghee for sale, while shop B has 50 tins of pure ghee and 60 tins of adulterated ghee. One tin of ghee is randomly purchased from one of the shops and found to be adulterated. What is the probability that it was purchased from shop B?

Solution:

Let's denote the following events:

E1 = The ghee is purchased from shop A

E2 = The ghee is purchased from shop B

A = The purchased ghee is adulterated

We can calculate the following probabilities:

P(E1) = ½ and P(E2) = ½

P(A|E1) = P(Purchasing adulterated ghee from shop A) = 40/70 = 4/7

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

Therefore,

\(\begin{array}{l}P(E_{2}|A)=\frac{P(A|E_{2})P(E_{2})}{P(A|E_{1})P(E_{1})+P(A|E_{2})P(E_{2})}\end{array} \)

\(\begin{array}{l}=\frac{6/11\times 1/2}{4/7 \times 1/2+6/11 \times 1/2}=\frac{21}{43}\end{array} \)

Hence, P(E2|A) = 21/43.

Practice Questions on Bayes’ Theorem

1. Selection and Probability of Change:
   If X, Y, and Z have chances of being selected as a manager in a company in the ratio 3:2:5, and the chances of introducing effective policy changes by them are 0.4, 0.6, and 0.7, respectively, then if a change is observed, what is the probability that it was due to the selection of Y?
2. Truth-teller and Dice Roll:
   A person tells the truth 3 out of 4 times. He rolls a die and says that the number shown is a five. What is the probability that the die actually shows five?
3. Drawing Red Balls from a Sack:
   A bag contains 5 balls. Two balls are drawn one after another without replacement and both are found to be red. What is the probability that all the balls in the bag are red?

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