Class 12 Maths Chapter 13: Probability

Probability is Chapter 13 of NCERT Class 12 Maths. It goes beyond Class 11 probability to cover conditional probability, the multiplication theorem, independent events, the theorem of total probability, Bayes' theorem, and random variables with their probability distributions and mean. Watch the five video lessons below, then revise with the notes, formulas and solved examples that follow.

What you will learn

Conditional probability P(A|B) and its properties
The multiplication theorem for two or more events
Independent events and how they differ from mutually exclusive events
The theorem of total probability and Bayes' theorem
Random variables, probability distributions and the mean (expectation)

Video Lessons

Probability - Part 1

Probability - Part 2

Probability - Part 3

Probability - Part 4

Probability - Part 5

Key Concepts and Formulas

Conditional Probability

Conditional probability answers the question: given that B has already happened, how likely is A? Knowing B shrinks the sample space to B itself, so P(B) must not be zero.

P(A|B) = P(A ∩ B)/P(B),  P(B) ≠ 0
P(S|B) = P(B|B) = 1
P(A′|B) = 1 − P(A|B)
P(A ∪ C | B) = P(A|B) + P(C|B) − P(A ∩ C | B)

Venn diagram of a die throw showing conditional probability P(E given F) = 2/3, Class 12 Maths Probability — Once F is known, only 2, 4 and 6 count

Multiplication Theorem

Rearranging the conditional probability formula gives the probability that two events both happen. This is the rule behind drawing items one after another without replacement.

P(A ∩ B) = P(A) P(B|A) = P(B) P(A|B)
P(A ∩ B ∩ C) = P(A) P(B|A) P(C|A ∩ B)

Independent Events

Two events are independent if the happening of one does not change the probability of the other. Independence is about probabilities, while mutual exclusivity is about outcomes not overlapping.

Independent: P(A ∩ B) = P(A) × P(B)
Equivalently: P(A|B) = P(A)
P(at least one) = 1 − P(A′) P(B′)

If A and B are independent, so are A and B′, A′ and B, and A′ and B′
Two events with non-zero probabilities cannot be both independent and mutually exclusive

Total Probability and Bayes' Theorem

If events E₁, E₂, ..., Eₙ form a partition of the sample space (they are pairwise disjoint, together cover S, and each has non-zero probability), then any event A can be broken up across them. Bayes' theorem reverses the conditioning and finds the probability of a cause after the result is known.

Total probability:
P(A) = Σ P(Eᵢ) P(A|Eᵢ)
Bayes' theorem:
P(Eᵢ|A) = P(Eᵢ) P(A|Eᵢ) / Σ P(Eⱼ) P(A|Eⱼ)

P(Eᵢ) are called prior probabilities and P(Eᵢ|A) are posterior probabilities

Random Variables and Their Mean

A random variable assigns a real number to each outcome of an experiment, for example the number of heads in three tosses. Its probability distribution lists every value with its probability.

Each pᵢ ≥ 0 and Σ pᵢ = 1
Mean: E(X) = μ = Σ xᵢ pᵢ

The mean is a weighted average and need not be one of the values of X

Solved Examples

Example 1: If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, find P(A|B) and check whether A and B are independent.

P(A|B) = 0.3/0.5 = 0.6
P(A) × P(B) = 0.6 × 0.5 = 0.3 = P(A ∩ B)
∴ P(A|B) = 0.6 and A, B are independent

Example 2: A die is thrown. Find the probability that the number is greater than 3, given that it is even.

E = {4, 5, 6}, F = {2, 4, 6}
E ∩ F = {4, 6}
P(E|F) = (2/6)/(3/6)
∴ P(E|F) = 2/3

Example 3: A bag has 5 red and 3 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are red.

P(1st red) = 5/8
P(2nd red | 1st red) = 4/7
P(both red) = 5/8 × 4/7 = 20/56
∴ P(both red) = 5/14

Example 4: A and B are independent with P(A) = 1/3 and P(B) = 1/4. Find P(A ∪ B).

P(A ∩ B) = 1/3 × 1/4 = 1/12
P(A ∪ B) = 1/3 + 1/4 − 1/12
= 4/12 + 3/12 − 1/12 = 6/12
∴ P(A ∪ B) = 1/2

Example 5: Bag I has 4 white and 2 black balls; Bag II has 3 white and 5 black balls. A bag is chosen at random and a white ball is drawn. Find the probability that it came from Bag I.

P(I) = P(II) = 1/2
P(W|I) = 4/6 = 2/3, P(W|II) = 3/8
P(W) = 1/2 × 2/3 + 1/2 × 3/8
     = 1/3 + 3/16 = 25/48
P(I|W) = (1/3) ÷ (25/48) = 48/75
∴ P(I|W) = 16/25

Tree diagram solving a two-bag problem with Bayes' theorem, Class 12 Maths Chapter 13 Probability — Tree diagram for Example 5: P(Bag I given white) = 16/25

Example 6: Machines A and B make 60% and 40% of a factory's bulbs, with 2% and 5% defective respectively. A bulb picked at random is defective. Find the probability that it was made by B.

P(D) = 0.6 × 0.02 + 0.4 × 0.05
     = 0.012 + 0.020 = 0.032
P(B|D) = 0.020/0.032 = 20/32
∴ P(B|D) = 5/8

Example 7: X takes values 0, 1, 2, 3 with probabilities k, 2k, 3k, 4k. Find k and E(X).

k + 2k + 3k + 4k = 1
10k = 1, so k = 1/10
E(X) = 0(1/10) + 1(2/10) + 2(3/10) + 3(4/10)
     = 0 + 2/10 + 6/10 + 12/10 = 20/10
∴ k = 1/10, E(X) = 2

Common Mistakes to Avoid

P(A|B) is not the same as P(B|A); read carefully which event is given and put it in the denominator.
Do not confuse independent with mutually exclusive; for mutually exclusive events P(A ∩ B) = 0, for independent events P(A ∩ B) = P(A)P(B).
In draws without replacement, both the numerator and the total reduce for the second draw.
In Bayes' theorem questions, first write the events E₁, E₂ and A in words and list all prior and conditional probabilities; this earns step marks in the CBSE board exam.

Quick Revision

P(A|B) = P(A ∩ B)/P(B)
P(A ∩ B) = P(A) P(B|A)
Independent: P(A ∩ B) = P(A) P(B)
Total: P(A) = Σ P(Eᵢ) P(A|Eᵢ)
Bayes: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ)/Σ P(Eⱼ)P(A|Eⱼ)
Distribution: Σ pᵢ = 1
Mean: E(X) = Σ xᵢ pᵢ