Binomial distribution: Class 12 practice
For n independent trials each with success probability p, P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Combinations
- Basic probability
The idea behind binomial distribution
For n independent trials each with success probability p, P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ. Mean = np and variance = np(1 − p).
Step-by-step method
- Identify n and p.
- Write P(X = r) with the formula.
- For 'at least', use 1 − P(fewer).
Worked example 1 (Basic)
A trial has success probability 1/4. In 3 independent trials, find P(exactly 0 successes).
- 3C0 (1/4)^0 (3/4)^3
- = 27/64
- Answer: 27/64
Worked example 2 (Standard)
A trial has success probability 1/3. In 5 independent trials, find P(exactly 5 successes).
- 5C5 (1/3)^5 (2/3)^0
- = 1/243
- Answer: 1/243
Worked example 3 (Challenge)
A fair coin is tossed 3 times. Find the mean and variance of the number of heads.
- Mean = np = 3/2
- Variance = np(1 − p) = 3/4
- Answer: Mean 3/2, variance 3/4
Common mistakes and misconceptions
- Forgetting the ⁿCᵣ factor
- Using p instead of 1 − p for failures
Practice questions
- (Basic) A trial has success probability 1/3. In 6 independent trials, find P(exactly 3 successes).
- (Basic) A trial has success probability 1/2. In 4 independent trials, find P(exactly 1 successes).
- (Standard) A trial has success probability 1/3. In 4 independent trials, find P(exactly 4 successes).
- (Standard) A trial has success probability 1/2. In 4 independent trials, find P(exactly 2 successes).
- (Challenge) A fair coin is tossed 5 times. Find the mean and variance of the number of heads.
- (Challenge) A fair coin is tossed 6 times. Find the mean and variance of the number of heads.
Answer key with full working
- Q1: 6C3 (1/3)^3 (2/3)^3 → = 160/729 → Answer: 160/729
- Q2: 4C1 (1/2)^1 (1/2)^3 → = 1/4 → Answer: 1/4
- Q3: 4C4 (1/3)^4 (2/3)^0 → = 1/81 → Answer: 1/81
- Q4: 4C2 (1/2)^2 (1/2)^2 → = 3/8 → Answer: 3/8
- Q5: Mean = np = 5/2 → Variance = np(1 − p) = 5/4 → Answer: Mean 5/2, variance 5/4
- Q6: Mean = np = 6/2 → Variance = np(1 − p) = 6/4 → Answer: Mean 3, variance 3/2
Syllabus note
This skill is listed under "Probability" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
- Conditional probability and Bayes' theorem
- Independent events and the multiplication rule
- Mean and variance of a random variable
- Law of total probability
- Derivatives of polynomials and slope of a tangent
- Definite integrals of polynomials
- Evaluating 2×2 and 3×3 determinants
- Matrix multiplication
- All Class 12 practice
Frequently asked questions
Which chapter is binomial distribution in?
"Probability", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.