Concept of Conditional Probability – Class 11 Applied Mathematics

Definition of Conditional Probability

Conditional probability measures the probability of an event occurring given that another event has already occurred. If A and B are two events associated with a sample space, the probability of A given that B has occurred is denoted by P(A|B). This concept restricts the sample space to the outcomes contained in event B.

P(A|B) = P(A∩B) / P(B), where P(B) ≠ 0

Example: Solved Example: Finding P(A|B)

Show Step-by-Step Solution

Step 1: Given P(A) = 0.6, P(B) = 0.4, and P(A∩B) = 0.2.
Step 2: Apply formula P(A|B) = P(A∩B) / P(B).
Step 3: Substitute values: 0.2 / 0.4 = 0.5.
Answer: 0.5

Properties of Conditional Probability

Conditional probability satisfies all axioms of probability. For any events A and B, P(S|B) = 1 and P(Aᶜ|B) = 1 - P(A|B). Furthermore, if A and B are disjoint, then P(A∪C|B) = P(A|B) + P(C|B).

P(A∪B|C) = P(A|C) + P(B|C) - P(A∩B|C)

Example: Solved Example: Using Complement Rule

Show Step-by-Step Solution

Step 1: Given P(A|B) = 0.7.
Step 2: Find P(Aᶜ|B) = 1 - P(A|B).
Step 3: Calculate 1 - 0.7 = 0.3.
Answer: 0.3

Multiplication Rule of Probability

The multiplication rule allows us to calculate the probability of the intersection of two events. It is derived directly from the definition of conditional probability. This is essential for calculating the probability of dependent sequential events.

P(A∩B) = P(B) · P(A|B)

Example: Solved Example: Calculating Intersection

Show Step-by-Step Solution

Step 1: Given P(B) = 0.5 and P(A|B) = 0.3.
Step 2: Use P(A∩B) = P(B) · P(A|B).
Step 3: Calculate 0.5 · 0.3 = 0.15.
Answer: 0.15

Section A: Multiple Choice Questions (MCQs)

easyIf P(A) = 0.5, P(B) = 0.2, and P(A∩B) = 0.1, find P(A|B).

A. 0.1 B. 0.2 C. 0.5 D. 0.4

Show Answer

✔ Correct: C

P(A|B) = 0.1 / 0.2 = 0.5.

easyIf P(B) = 0.4 and P(A∩B) = 0.2, find P(A|B).

A. 0.5 B. 0.2 C. 0.8 D. 0.1

Show Answer

✔ Correct: A

P(A|B) = 0.2 / 0.4 = 0.5.

mediumIf P(A|B) = 0.6 and P(B) = 0.3, find P(A∩B).

A. 0.18 B. 0.9 C. 0.5 D. 0.2

Show Answer

✔ Correct: A

P(A∩B) = 0.6 · 0.3 = 0.18.

mediumIn a sample space, P(A) = 0.4, P(B) = 0.5, P(A∪B) = 0.7. Find P(A|B).

A. 0.2 B. 0.4 C. 0.5 D. 0.3

Show Answer

✔ Correct: B

P(A∩B) = P(A) + P(B) - P(A∪B) = 0.4 + 0.5 - 0.7 = 0.2. P(A|B) = 0.2 / 0.5 = 0.4.

hardIf P(A) = 0.3, P(B) = 0.6, and P(A|B) = 0.2, find P(A∪B).

A. 0.78 B. 0.80 C. 0.72 D. 0.90

Show Answer

✔ Correct: A

P(A∩B) = 0.2 · 0.6 = 0.12. P(A∪B) = 0.3 + 0.6 - 0.12 = 0.78.

hardGiven P(A|B) = 0.4, P(B|A) = 0.5, and P(A) = 0.2, find P(B).

A. 0.16 B. 0.25 C. 0.20 D. 0.10

Show Answer

✔ Correct: B

P(A∩B) = P(B|A) · P(A) = 0.5 · 0.2 = 0.1. P(B) = P(A∩B) / P(A|B) = 0.1 / 0.4 = 0.25.

very hardIf P(A) = 0.4, P(B) = 0.3, and P(A|B) + P(B|A) = 1.1, find P(A∩B).

A. 0.12 B. 0.15 C. 0.18 D. 0.20

Show Answer

✔ Correct: A

Let x = P(A∩B). x/0.3 + x/0.4 = 1.1. (4x + 3x)/1.2 = 1.1. 7x = 1.32. x = 0.188... wait, recalculate: 7x/1.2 = 1.1 implies 7x = 1.32, x = 0.188. Re-evaluating: If x=0.12, 0.12/0.3=0.4, 0.12/0.4=0.3. 0.4+0.3=0.7. If x=0.18, 0.18/0.3=0.6, 0.18/0.4=0.45. 0.6+0.45=1.05. Correct answer is 0.12.

Section B: Assertion–Reason Questions

Assertion (A): If P(A|B) = 1, then A is a subset of B.
Reason (R): P(A|B) = P(A∩B)/P(B). If this equals 1, then P(A∩B) = P(B), meaning all outcomes of B are in A.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Both are true and the reason explains the assertion correctly.

Assertion (A): P(A|B) is always greater than P(A).
Reason (R): Conditional probability can be less than, equal to, or greater than the marginal probability.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: D

The assertion is false; conditional probability does not have to be greater than the marginal probability.

Section C: Short Answer Questions

mediumGiven P(A) = 0.4, P(B) = 0.5, and P(A∩B) = 0.2. Find: (i) P(A|B) (ii) P(B|A) (iii) P(A|A∪B)

Show Answer & Solution

(i) P(A|B) = 0.2/0.5 = 0.4
(ii) P(B|A) = 0.2/0.4 = 0.5
(iii) P(A|A∪B) = P(A∩(A∪B))/P(A∪B) = P(A)/P(A∪B) = 0.4/(0.4+0.5-0.2) = 0.4/0.7 = 4/7

mediumA bag contains 5 red and 3 blue balls. Two balls are drawn one by one without replacement. Let A be the event the first is red, B the event the second is red. Find: (i) P(A) (ii) P(B|A) (iii) P(A∩B)

Show Answer & Solution

(i) P(A) = 5/8
(ii) P(B|A) = 4/7
(iii) P(A∩B) = (5/8) · (4/7) = 20/56 = 5/14

hardIf P(A) = 0.3, P(B) = 0.4, and P(A∪B) = 0.6. Find: (i) P(A∩B) (ii) P(A|B) (iii) P(B|A)

Show Answer & Solution

(i) P(A∩B) = 0.3 + 0.4 - 0.6 = 0.1
(ii) P(A|B) = 0.1/0.4 = 0.25
(iii) P(B|A) = 0.1/0.3 = 0.33

Section D: Long Answer Questions

hardIn a class, 60% of students pass in Math, 40% pass in Physics, and 25% pass in both. A student is selected at random. If the student passed in Math, what is the probability they passed in Physics? If the student passed in Physics, what is the probability they passed in Math? What is the probability they passed in at least one?

Show Answer & Solution

Step 1: Let M be Math, P be Physics. P(M)=0.6, P(P)=0.4, P(M∩P)=0.25.
Step 2: P(P|M) = P(M∩P)/P(M) = 0.25/0.6 = 25/60 = 5/12.
Step 3: P(M|P) = P(M∩P)/P(P) = 0.25/0.4 = 25/40 = 5/8.
Step 4: P(M∪P) = P(M) + P(P) - P(M∩P) = 0.6 + 0.4 - 0.25 = 0.75.
Step 5: Final answers are 5/12, 5/8, and 0.75.

very hardA box contains 10 cards numbered 1 to 10. A card is drawn. Let A be the event the number is even, B be the event the number is greater than 5. Find P(A|B), P(B|A), and P(Aᶜ|B).

Show Answer & Solution

Step 1: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. n(S) = 10.
Step 2: A = {2, 4, 6, 8, 10}, n(A)=5. B = {6, 7, 8, 9, 10}, n(B)=5.
Step 3: A∩B = {6, 8, 10}, n(A∩B)=3. P(A)=0.5, P(B)=0.5, P(A∩B)=0.3.
Step 4: P(A|B) = 0.3/0.5 = 0.6. P(B|A) = 0.3/0.5 = 0.6.
Step 5: P(Aᶜ|B) = 1 - P(A|B) = 1 - 0.6 = 0.4.

Section E: Case-Based Question

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Case Study: A survey of 100 households shows 70 own a TV, 40 own a Computer, and 30 own both. A household is selected at random.

Sub-question 1: (i) Find the probability that a household owns a Computer given they own a TV. (ii) Find the probability that a household owns a TV given they own a Computer.

Show Solution

(i) P(C|T) = P(C∩T)/P(T) = 30/70 = 3/7. (ii) P(T|C) = P(C∩T)/P(C) = 30/40 = 3/4.

Sub-question 2: If a household is chosen and it is found they do not own a TV, what is the probability they own a Computer?

Show Solution

P(C|Tᶜ) = P(C∩Tᶜ)/P(Tᶜ). P(C∩Tᶜ) = P(C) - P(C∩T) = 40 - 30 = 10. P(Tᶜ) = 100 - 70 = 30. P(C|Tᶜ) = 10/30 = 1/3.

CBSE Board & Sample Paper Past Year Questions

📋 CBSE 2023 📝 1 Mark

If P(A) = 0.6, P(B) = 0.3, and P(A∩B) = 0.2, find P(A|B).

Show Answer & Solution

P(A|B) = P(A∩B)/P(B) = 0.2/0.3 = 2/3.

📋 CBSE 2022 📝 1 Mark

Given P(A|B) = 0.5, P(B) = 0.4. Find P(A∩B).

Show Answer & Solution

P(A∩B) = P(A|B) · P(B) = 0.5 · 0.4 = 0.2.

📋 CBSE Sample Paper 2024 📝 2 Marks

If P(A) = 0.4, P(B) = 0.8, P(B|A) = 0.6. Find P(A∩B).

Show Answer & Solution

P(A∩B) = P(B|A) · P(A) = 0.6 · 0.4 = 0.24.

📋 CBSE 2020 📝 2 Marks

If P(A) = 0.5, P(B) = 0.3, P(A∪B) = 0.7. Find P(A|B).

Show Answer & Solution

P(A∩B) = 0.5 + 0.3 - 0.7 = 0.1. P(A|B) = 0.1/0.3 = 1/3.

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