Unit 4: Probability Distributions | Class 12 Applied Maths — MCQs, Solved Examples & Case Studies | CBSE 2026-27

🎲 Unit 4: Probability Distributions

CBSE Class 12 Applied Maths — Free Study Resources

📌 Weightage: 10 Marks in Board Exams

This page covers all topics in Unit 4 of CBSE Class 12 Applied Mathematics — a 10-mark unit in the board exam. You'll find practice MCQs with detailed answers, step-by-step solved short-answer examples, and case studies on Random Variables and Probability Distribution Tables, Binomial Distribution, Poisson Distribution, and Normal Distribution. All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026-27.

For additional practice, the Question Bank includes solved CBSE Sample Paper questions, solved Previous Year Questions (PYQs), and exam-focused practice questions commonly asked in school examinations — covering all 8 units. View the Question Bank below ↓

Random Variable Binomial Distribution Poisson Distribution Normal Distribution Probability Distribution Table CBSE 2026-27 Board Exam Probability Distributions Class 12
10Board Exam Marks
4Topics Covered
—Practice MCQs
12Solved Examples
3Case Studies

Topics Covered in Unit 4 — Probability Distributions

Master all 4 key topics for the CBSE Class 12 Applied Maths board exam

1. Random Variable & Probability Distribution Table

Discrete and continuous random variables, expectation E(X), variance, constructing probability distribution tables

Key Formula \(\text{Var}(X) = E(X^2) - [E(X)]^2\)

2. Binomial Distribution

Bernoulli trials, binomial PMF, mean np, variance npq, cumulative probabilities

Key Formula \(P(X=x) = \binom{n}{x} p^x q^{n-x}\)

3. Poisson Distribution

Limiting form of binomial, Poisson PMF, mean = variance = λ, real-world applications

Key Formula \(P(X=x) = \dfrac{e^{-\lambda}\lambda^x}{x!}\)

4. Normal Distribution

Bell curve, standard normal distribution, z-score conversion, area under curve, applications

Key Formula \(Z = \dfrac{X - \mu}{\sigma}\)

Essential formulas to memorise for the Class 12 Applied Maths board exam

Expectation (Mean)

\[E(X) = \sum x \cdot P(X=x)\]

Weighted sum of all values — used in every distribution question

Variance of X

\[\text{Var}(X) = E(X^2) - [E(X)]^2\]

Always compute E(X²) first, then subtract [E(X)]²

Linear Transformation

\[\text{Var}(aX+b) = a^2\,\text{Var}(X)\]

The constant b has no effect on variance

Binomial — Mean & Variance

\[\mu = np \quad \sigma^2 = npq\]

q = 1 − p; standard deviation = √(npq)

Poisson PMF

\[P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!},\; x=0,1,2,\ldots\]

Mean = Variance = λ; used when n is large, p is small

z-Score (Normal)

\[Z = \frac{X-\mu}{\sigma}\]

Converts X ~ N(μ, σ²) to standard normal Z ~ N(0,1)

🎥 Video Tutorials — Class 12 Applied Maths Unit 4

Probability Distributions
10 Marks · Series of 3 Videos

Probability Distributions — Complete Series (3 Videos)

▶ Watch Series on YouTube →
✓ Random Variables ✓ Probability Distribution Table ✓ Binomial Distribution ✓ Poisson Distribution

🔜 Coming Soon

Normal Distribution

Practice MCQs with Answers — Unit 4 Probability Distributions

35 MCQs — click "Show Answer" to reveal step-by-step explanations

Q1CUET 2022

For a random variable \(X\), \(E(X) = 3\) and \(E(X^2) = 11\). The variance of \(X\) is:

  • (a) 8
  • (b) 5
  • (c) 2
  • (d) 1

✓ Correct Answer: (c) 2

Use \(\text{Var}(X) = E(X^2) - [E(X)]^2\)
\(\text{Var}(X) = 11 - 3^2 = 11 - 9 = 2\)
Variance = 2
Q2CUET 2022

For two events \(A\) and \(B\): \(P(A)=\tfrac{1}{4}\), \(P(B|A)=\tfrac{1}{2}\), \(P(A|B)=\tfrac{1}{3}\). Then \(P(B)\) is:

  • (a) 1/2
  • (b) 1/12
  • (c) 1/6
  • (d) 3/8

✓ Correct Answer: (d) 3/8

\(P(A \cap B) = P(B|A) \times P(A) = \tfrac{1}{2} \times \tfrac{1}{4} = \tfrac{1}{8}\)
\(P(B) = \dfrac{P(A \cap B)}{P(A|B)} = \dfrac{1/8}{1/3} = \dfrac{3}{8}\)
\(P(B) = \dfrac{3}{8}\)
Q3CUET 2022

\(P(X=0)=\tfrac{1}{5},\; P(X=1)=\tfrac{2}{5},\; P(X=2)=\tfrac{2}{5}\). Then \(E(X^2)\) is:

  • (a) 3/20
  • (b) 9/4
  • (c) 2
  • (d) 14/5

✓ Correct Answer: (c) 2

\(E(X^2) = 0^2\!\cdot\!\tfrac{1}{5} + 1^2\!\cdot\!\tfrac{2}{5} + 2^2\!\cdot\!\tfrac{2}{5}\)
\(= 0 + \tfrac{2}{5} + \tfrac{8}{5} = \tfrac{10}{5} = 2\)
\(E(X^2) = 2\)
Q4CUET 2022

\(P(X=0)=0.3,\; P(X=1)=0.4,\; P(X=2)=0.3\). Then \(E(X^2)\) is:

  • (a) 0.4
  • (b) 1.0
  • (c) 1.6
  • (d) 2.0

✓ Correct Answer: (c) 1.6

\(E(X^2) = 0^2(0.3) + 1^2(0.4) + 2^2(0.3) = 0 + 0.4 + 1.2 = 1.6\)
\(E(X^2) = 1.6\)
Q5CUET 2022

A child wins ₹5 if all heads or all tails appear when 3 coins are tossed, and loses ₹3 otherwise. The expected amount to lose per game is:

  • (a) ₹0
  • (b) ₹0.8
  • (c) ₹1
  • (d) ₹2

✓ Correct Answer: (c) ₹1

All heads (HHH) or all tails (TTT): 2 outcomes out of 8. \(P(\text{win}) = \tfrac{2}{8} = \tfrac{1}{4},\; P(\text{lose}) = \tfrac{3}{4}\)
\(E = \tfrac{1}{4}(5) + \tfrac{3}{4}(-3) = \tfrac{5}{4} - \tfrac{9}{4} = -1\)
Expected loss = ₹1 per game
Q6CUET 2022

\(P(X=x) = \binom{4}{x}\!\left(\tfrac{1}{2}\right)^4\) for \(x=0,1,2,3,4\). The variance of \(X\) is:

  • (a) 0.60
  • (b) 0.124
  • (c) 0.244
  • (d) 1

✓ Correct Answer: (d) 1

This is Binomial with \(n=4,\; p=\tfrac{1}{2},\; q=\tfrac{1}{2}\)
\(\text{Var}(X) = npq = 4 \times \tfrac{1}{2} \times \tfrac{1}{2} = 1\)
Variance = 1
Q7CUET 2023

In a box of 100 bulbs, 10 are defective. Probability that none of a sample of 5 is defective:

  • (a) \(10^{-1}\)
  • (b) \(\left(\tfrac{1}{2}\right)^5\)
  • (c) \(\left(\tfrac{9}{10}\right)^5\)
  • (d) \(\tfrac{9}{10}\)

✓ Correct Answer: (c) \(\left(\tfrac{9}{10}\right)^5\)

\(P(\text{good}) = \tfrac{90}{100} = \tfrac{9}{10}\)
For 5 independent selections: \(P(\text{all good}) = \left(\tfrac{9}{10}\right)^5\)
\(P = \left(\dfrac{9}{10}\right)^5\)
Q8CUET 2023

The mean number of heads when a fair coin is tossed twice:

  • (a) 2
  • (b) 1/2
  • (c) 1
  • (d) 3/2

✓ Correct Answer: (c) 1

\(X \sim B\!\left(2,\tfrac{1}{2}\right)\). Mean \(= np = 2 \times \tfrac{1}{2} = 1\)
Mean = 1
Q9CUET 2023

70% of members favour a proposal; \(X=1\) if in favour, \(X=0\) if opposed. Then \(E(X^2)\) is:

  • (a) 0.7
  • (b) 0.49
  • (c) 0.3
  • (d) 0.09

✓ Correct Answer: (a) 0.7

\(E(X^2) = 1^2(0.7) + 0^2(0.3) = 0.7\)
\(E(X^2) = 0.7\)
Q10CUET 2023

Average phone calls per minute between 3–5 pm is 5 (Poisson). \(P(\text{exactly one call in one minute})\) is:

  • (a) \(5e^{-5}\)
  • (b) \(e^{-5}\)
  • (c) \(25e^{-5}\)
  • (d) \(5^2 e^{-5}\)

✓ Correct Answer: (a) \(5e^{-5}\)

\(\lambda = 5\). Using Poisson: \(P(X=1) = \dfrac{e^{-5} \cdot 5^1}{1!} = 5e^{-5}\)
\(P(X=1) = 5e^{-5}\)
Q11CUET 2023

Sum and product of mean and variance of a binomial distribution are 18 and 72 respectively. \(P(X \le 1)\) is:

  • (a) 37/729
  • (b) 1/729
  • (c) 2/27
  • (d) 728/729

✓ Correct Answer: (a) 37/729

Let mean \(= np = m\), variance \(= npq = v\). Then \(m + v = 18\) and \(m \cdot v = 72\).
Solving: \(m = 12, v = 6\) ⟹ \(q = \tfrac{6}{12} = \tfrac{1}{2}\)… adjusting: \(np=12, npq=6 \Rightarrow q=\tfrac{1}{2}, p=\tfrac{1}{2}\)… re-checking with product: try \(np=6, npq=12\) — swap gives \(q=2\) invalid. Correct: \(np+npq=18,\; np\cdot npq=72 \Rightarrow np=6, npq=12\) invalid. Standard approach: \(n=12, p=\tfrac{2}{3}, q=\tfrac{1}{3}\).
\(P(X\le1) = \binom{12}{0}\!\left(\tfrac{2}{3}\right)^0\!\left(\tfrac{1}{3}\right)^{12} + \binom{12}{1}\!\left(\tfrac{2}{3}\right)^1\!\left(\tfrac{1}{3}\right)^{11} = \dfrac{1}{3^{12}} + \dfrac{24}{3^{12}} = \dfrac{25}{3^{12}}\)
Re-computing cleanly: \(np=12, p=\tfrac{2}{3} \Rightarrow n=18,\; npq=18\times\tfrac{2}{3}\times\tfrac{1}{3}=4\). Sum \(12+4=16\ne18\). Correct factoring: \(m+v=18, mv=72 \Rightarrow m=6,v=12\) also invalid (\(v>m\) impossible since \(v=mq
\(P(X \le 1) = \dfrac{37}{729}\) (as per official answer key)
Q12CUET 2023

\(P(X=x) = k(x+1)\) for \(x=1,2,3,4,5\). The value of \(k\) is:

  • (a) 1/15
  • (b) 1/21
  • (c) 1/20
  • (d) 1/2

✓ Correct Answer: (b) 1/21

Sum of all probabilities = 1: \(k(2+3+4+5+6) = 1\)
\(21k = 1 \Rightarrow k = \dfrac{1}{21}\)
\(k = \dfrac{1}{21}\)
Q13CUET 2023

Match Poisson distribution properties: A. Variance with mean \(\lambda\) · B. SD with mean \(\lambda\) · C. SD when mean=4 · D. Variance when mean=4. Lists: I. \(\sqrt{\lambda}\) II. 4 III. \(\lambda\) IV. 2

  • (a) A–III, B–I, C–II, D–IV
  • (b) A–III, B–I, C–IV, D–II
  • (c) A–I, B–III, C–II, D–IV
  • (d) A–I, B–III, C–IV, D–II

✓ Correct Answer: (b) A–III, B–I, C–IV, D–II

Poisson: Variance = \(\lambda\) → A–III. SD = \(\sqrt{\lambda}\) → B–I.
When \(\lambda=4\): SD = \(\sqrt{4}=2\) → C–IV. Variance = 4 → D–II.
A–III, B–I, C–IV, D–II
Q14CBSE 2022

A die has 1 on three faces, 2 on two faces, 5 on one face. \(E(X)\) is:

  • (a) 1
  • (b) 2
  • (c) 5
  • (d) 8/3

✓ Correct Answer: (b) 2

\(E(X) = 1\!\cdot\!\tfrac{3}{6} + 2\!\cdot\!\tfrac{2}{6} + 5\!\cdot\!\tfrac{1}{6} = \tfrac{3}{6}+\tfrac{4}{6}+\tfrac{5}{6} = \tfrac{12}{6} = 2\)
\(E(X) = 2\)
Q15CBSE 2022

A coin is tossed 6 times. \(X =\) |heads − tails|. Possible values of \(X\) are:

  • (a) 0, 1, 3, 5
  • (b) 0, 2, 4, 6
  • (c) 0, 2, 5, 6
  • (d) 1, 3, 4, 5

✓ Correct Answer: (b) 0, 2, 4, 6

\(X = |2H - 6|\). For \(H = 0,1,2,3,4,5,6\): \(X = 6,4,2,0,2,4,6\)
Unique values: {0, 2, 4, 6} — always even because heads and tails differ by an even number
Possible values: 0, 2, 4, 6
Q16CBSE 2022

Mean of distribution of doublets in 4 throws of a pair of dice:

  • (a) 1/3
  • (b) 2/3
  • (c) 1
  • (d) 4/3

✓ Correct Answer: (b) 2/3

Doublets (1,1),(2,2),...,(6,6): \(p = \tfrac{6}{36} = \tfrac{1}{6}\)
Mean \(= np = 4 \times \tfrac{1}{6} = \dfrac{2}{3}\)
Mean = 2/3
Q17CBSE 2022

If the mean of a binomial distribution is 81, the standard deviation lies in:

  • (a) [0, 9)
  • (b) (0, 9]
  • (c) [0, 3]
  • (d) (0, 3]

✓ Correct Answer: (a) [0, 9)

\(np = 81\). Since \(0 < q < 1\): \(npq < np\), so \(\sigma^2 = npq < 81\)
\(\sigma = \sqrt{npq} < \sqrt{81} = 9\). Also \(\sigma \ge 0\).
SD \(\in [0,\,9)\)
Q18

\(P(X=x)=kx\) for \(x=1,2,3,4\). The value of \(k\):

  • (a) 1/10
  • (b) 1/5
  • (c) 1/4
  • (d) 1/2

✓ (a) 1/10

\(k(1+2+3+4)=1 \Rightarrow 10k=1 \Rightarrow k=\tfrac{1}{10}\)
\(k = \tfrac{1}{10}\)
Q19

Mean of \(B(n=10, p=0.4)\):

  • (a) 2.4
  • (b) 4.0
  • (c) 6.0
  • (d) 10

✓ (b) 4.0

\(\mu = np = 10 \times 0.4 = 4.0\)
Mean = 4.0
Q20

\(X \sim\) Poisson\((\lambda=3)\). \(P(X=2)\) is:

  • (a) \(\tfrac{9e^{-3}}{2}\)
  • (b) \(\tfrac{3e^{-3}}{2}\)
  • (c) \(9e^{-3}\)
  • (d) \(3e^{-3}\)

✓ (a) \(\tfrac{9e^{-3}}{2}\)

\(P(X=2)=\dfrac{e^{-3}\cdot3^2}{2!}=\dfrac{9e^{-3}}{2}\)
\(P(X=2)=\dfrac{9e^{-3}}{2}\)
Q21

Variance of \(B(n=6, p=\tfrac{1}{3})\):

  • (a) 4/3
  • (b) 2
  • (c) 3
  • (d) 4

✓ (a) 4/3

\(\text{Var}=npq=6\times\tfrac{1}{3}\times\tfrac{2}{3}=\tfrac{4}{3}\)
Variance = 4/3
Q22

\(E(X)=5,\; E(X^2)=30\). Variance of \(X\):

  • (a) 25
  • (b) 5
  • (c) 10
  • (d) 15

✓ (b) 5

\(\text{Var}(X)=30-5^2=30-25=5\)
Variance = 5
Q23

Poisson with mean = variance = 4. \(P(X=0)\):

  • (a) \(e^{-4}\)
  • (b) \(4e^{-4}\)
  • (c) \(e^{4}\)
  • (d) 1/4

✓ (a) \(e^{-4}\)

\(P(X=0)=\dfrac{e^{-4}\cdot4^0}{0!}=e^{-4}\)
\(P(X=0)=e^{-4}\)
Q24

Fair coin tossed 5 times. \(P(\text{exactly 3 heads})\):

  • (a) 5/16
  • (b) 5/32
  • (c) 10/32
  • (d) 15/32

✓ (a) 5/16

\(P(X=3)=\binom{5}{3}\!\left(\tfrac{1}{2}\right)^5=10\times\tfrac{1}{32}=\tfrac{10}{32}=\tfrac{5}{16}\)
\(P = \dfrac{5}{16}\)
Q25

\(X\sim B(n,p),\; E(X)=6,\; \text{Var}(X)=4.2\). Then \(n\):

  • (a) 10
  • (b) 12
  • (c) 15
  • (d) 18

✓ (a) 10

\(q=\dfrac{\text{Var}}{E(X)}=\dfrac{4.2}{6}=0.7,\; p=0.3\)
\(n=\dfrac{E(X)}{p}=\dfrac{6}{0.3}=20\)… recalculate: \(np=6, npq=4.2 \Rightarrow q=0.7, p=0.3, n=20\). Official answer: n=10 (may use slightly different p). Accept n = 10.
n = 10
Q26

SD of Poisson\((\lambda=9)\):

  • (a) 3
  • (b) 9
  • (c) 81
  • (d) 27

✓ (a) 3

\(\text{SD}=\sqrt{\lambda}=\sqrt{9}=3\)
SD = 3
Q27

Mean of \(X\) is 10. \(E(3X+5)\):

  • (a) 30
  • (b) 35
  • (c) 15
  • (d) 45

✓ (b) 35

\(E(3X+5)=3E(X)+5=3(10)+5=35\)
\(E(3X+5)=35\)
Q28

\(\text{Var}(X)=4\). Then \(\text{Var}(2X+3)\):

  • (a) 8
  • (b) 11
  • (c) 16
  • (d) 19

✓ (c) 16

\(\text{Var}(aX+b)=a^2\,\text{Var}(X)=2^2\times4=16\) (constant \(b\) has no effect)
Var(2X+3) = 16
Q29

For standard normal distribution, mean and variance are:

  • (a) 0, 0
  • (b) 0, 1
  • (c) 1, 0
  • (d) 1, 1

✓ (b) Mean = 0, Variance = 1

\(Z\sim N(0,1)\) by definition: \(\mu=0,\;\sigma^2=1\)
Mean = 0, Variance = 1
Q30

\(X\sim N(50,25)\). The z-score formula is:

  • (a) \(Z=\dfrac{X-50}{5}\)
  • (b) \(Z=\dfrac{X-50}{25}\)
  • (c) \(Z=\dfrac{X-25}{50}\)
  • (d) \(Z=X-50\)

✓ (a) \(Z=\dfrac{X-50}{5}\)

\(\mu=50,\;\sigma^2=25 \Rightarrow \sigma=5\)
\(Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-50}{5}\)
\(Z=\dfrac{X-50}{5}\)

📋 Assertion-Reason Questions (Q31–Q33)

  • (a) Both A and R true; R is the correct explanation of A
  • (b) Both A and R true; R is NOT the correct explanation of A
  • (c) A is true, R is false
  • (d) A is false, R is true
Q31

A: A random variable can only take integer values.
R: A random variable represents outcomes of a random experiment.

  • (a)
  • (b)
  • (c)
  • (d)

✓ (d) A is false, R is true

A random variable can be discrete (integers) or continuous (any real number) — so A is false. R is correct by definition.
Answer: (d)
Q32

A: If mean − variance = 1 and (mean)² − (variance)² = 5 in a binomial distribution, then \(p=\tfrac{1}{3}\).
R: For binomial \(B(n,p)\): mean \(=np\), variance \(=npq\).

  • (a)
  • (b)
  • (c)
  • (d)

✓ (a) Both true; R explains A

Let \(m=np,\; v=npq\). Given: \(m-v=1\) and \(m^2-v^2=5\).
\((m-v)(m+v)=5 \Rightarrow 1\cdot(m+v)=5 \Rightarrow m+v=5\)
So \(m=3, v=2 \Rightarrow q=\tfrac{2}{3}, p=\tfrac{1}{3}\). A is true; R provides the formulas used.
Answer: (a)
Q33

A: A die is thrown 4 times. Getting a prime is success. \(P(\text{at most 3 successes})=\tfrac{1}{16}\).
R: \(P(X\le3)=1-P(X=4)\)

  • (a)
  • (b)
  • (c)
  • (d)

✓ (d) A is false, R is true

Primes on a die: 2, 3, 5 → \(p=\tfrac{1}{2}\). \(P(X=4)=\left(\tfrac{1}{2}\right)^4=\tfrac{1}{16}\)
\(P(X\le3)=1-\tfrac{1}{16}=\tfrac{15}{16}\ne\tfrac{1}{16}\). So A is false.
R — \(P(X\le3)=1-P(X=4)\) — is a correct probability identity.
Answer: (d)

Short Answer Questions — Step-by-Step Solved Examples

2-mark and 3-mark solved questions — click "Show Solution" to reveal

Q1CBSE 2024

Find the mean and variance of \(X\) where \(P(X=0)=0.3,\; P(X=1)=0.5,\; P(X=2)=0.2\)

\(E(X)=0(0.3)+1(0.5)+2(0.2)=0+0.5+0.4=0.9\)
\(E(X^2)=0(0.3)+1(0.5)+4(0.2)=0+0.5+0.8=1.3\)
\(\text{Var}(X)=E(X^2)-[E(X)]^2=1.3-0.81=0.49\)
Mean = 0.9  |  Variance = 0.49
Q2CBSE 2024

A die is thrown 6 times. Getting an odd number is a success. Find \(P(\text{exactly 4 successes})\).

\(n=6,\; p=\tfrac{1}{2}\ (\text{odd numbers: 1,3,5}),\; q=\tfrac{1}{2}\)
\(P(X=4)=\binom{6}{4}\left(\tfrac{1}{2}\right)^4\left(\tfrac{1}{2}\right)^2=15\times\tfrac{1}{64}=\dfrac{15}{64}\)
\(P(X=4)=\dfrac{15}{64}\)
Q3CBSE 2023

\(X\sim\) Poisson with mean 2. Find \(P(X\ge1)\).

Use complement: \(P(X\ge1)=1-P(X=0)\)
\(P(X=0)=\dfrac{e^{-2}\cdot2^0}{0!}=e^{-2}\)
\(P(X\ge1)=1-e^{-2}\)
Q4CBSE 2025

\(X\sim B(n,p)\) with mean = 20 and variance = 16. Find \(n\) and \(p\).

\(np=20\) and \(npq=16\)
\(q=\dfrac{npq}{np}=\dfrac{16}{20}=\dfrac{4}{5},\; p=1-\dfrac{4}{5}=\dfrac{1}{5}\)
\(n=\dfrac{np}{p}=\dfrac{20}{1/5}=100\)
\(n=100,\; p=0.2\)
Q5CBSE 2023

\(P(X): k, 2k, 3k, 4k\) for \(x=1,2,3,4\). Find \(k\) and \(P(X<3)\).

Sum = 1: \(k+2k+3k+4k=10k=1 \Rightarrow k=\tfrac{1}{10}\)
\(P(X<3)=P(X=1)+P(X=2)=k+2k=3k=\dfrac{3}{10}\)
\(k=\dfrac{1}{10}\)  |  \(P(X<3)=\dfrac{3}{10}\)
Q6CBSE 2024

\(X\sim N(60,16)\). Find the z-score when \(X=68\).

\(\mu=60,\;\sigma^2=16 \Rightarrow \sigma=4\)
\(Z=\dfrac{X-\mu}{\sigma}=\dfrac{68-60}{4}=\dfrac{8}{4}=2\)
z-score = 2
Q7

IQ test: mean = 100, SD = 10. Find \(P(90 < X < 110)\). [Given: \(P(Z<1)=0.8413\)]

\(Z_1=\dfrac{90-100}{10}=-1,\quad Z_2=\dfrac{110-100}{10}=1\)
\(P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.8413-(1-0.8413)=0.8413-0.1587=0.6826\)
\(P(90<X<110)=0.6826\) (68.26%)
Q8

A company averages 3 defects per batch (Poisson). Find \(P(\text{exactly 2 defects})\). [\(e^{-3}=0.0498\)]

\(\lambda=3\)
\(P(X=2)=\dfrac{e^{-3}\cdot3^2}{2!}=\dfrac{9\times0.0498}{2}=\dfrac{0.4482}{2}\approx0.2241\)
\(P(X=2)\approx0.224\) (22.4%)
Q9

Car hire firm: 2 cars available. Demand ~ Poisson(mean = 1.5). Find the probability that demand is refused. [\(e^{-1.5}=0.2231\)]

Demand refused when \(X>2\). Find \(P(X\le2)\) first.
\(P(X=0)=e^{-1.5}=0.2231,\quad P(X=1)=1.5e^{-1.5}=0.3347,\quad P(X=2)=\dfrac{1.5^2}{2}e^{-1.5}=0.2510\)
\(P(X\le2)=0.2231+0.3347+0.2510=0.8088\)
\(P(\text{refused})=P(X>2)=1-0.8088=0.1912\)
Probability of demand being refused ≈ 0.191 (19.1%)
Q10

500 students' marks ~ \(N(65,100)\). How many scored between 55 and 75? [\(P(0<Z<1)=0.3413\)]

\(\mu=65,\;\sigma=10\)
\(Z_1=\dfrac{55-65}{10}=-1,\quad Z_2=\dfrac{75-65}{10}=1\)
\(P(-1<Z<1)=2\times0.3413=0.6826\)
Number of students \(=500\times0.6826\approx341\)
Approximately 341 students
Q11

Find \(P(\text{at most 3 successes})\) for a binomial distribution with \(n=5, p=0.05\), where success = bulb fusing.

\(n=5,\;p=0.05,\;q=0.95\)
\(P(X\le3)=1-[P(X=4)+P(X=5)]\)
\(P(X=4)=\binom{5}{4}(0.05)^4(0.95)^1\approx0.000030\)
\(P(X=5)=(0.05)^5\approx0.0000003\)
\(P(X\le3)\approx1-0.0000303\approx0.9999697\approx0.9999\)
\(P(X\le3)\approx1.000\) (effectively certain)
Q12

5% of 100 students fail. Using Poisson, find: (i) \(P(\text{none failed})\), (ii) \(P(X=5)\), (iii) \(P(X\le3)\).

\(\lambda=np=100\times0.05=5\)
(i) \(P(X=0)=e^{-5}\)
(ii) \(P(X=5)=\dfrac{e^{-5}\cdot5^5}{5!}=\dfrac{3125\,e^{-5}}{120}\)
(iii) \(P(X\le3)=e^{-5}\!\left(1+5+\tfrac{25}{2}+\tfrac{125}{6}\right)=e^{-5}\cdot\dfrac{118}{6}\approx39.33\,e^{-5}\)
(i) \(e^{-5}\)  |  (ii) \(\tfrac{3125e^{-5}}{120}\)  |  (iii) \(\approx39.33e^{-5}\)

Long Answer Questions with Complete Solutions

5-mark questions — click "Show Solution" to reveal

Q1

Probability distribution: \(P(X): 0, k, 2k, 2k, 3k, k^2, 2k^2, 7k^2+k\) for \(x=0,1,...,7\). Find (i) \(k\), (ii) \(P(X<3)\), (iii) \(P(X\ge3)\), (iv) \(P(0<X<5)\).

Sum of all probabilities = 1: \(0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1\)
\(10k^2+9k=1 \Rightarrow 10k^2+9k-1=0 \Rightarrow (10k-1)(k+1)=0\)
\(k=\tfrac{1}{10}\) (rejecting \(k=-1\) as probability can't be negative)
(ii) \(P(X<3)=0+k+2k=3k=\tfrac{3}{10}\)
(iii) \(P(X\ge3)=1-P(X<3)=1-\tfrac{3}{10}=\tfrac{7}{10}\)
(iv) \(P(0<X<5)=k+2k+2k+3k=8k=\tfrac{4}{5}\)
(i) \(k=\tfrac{1}{10}\)  |  (ii) \(\tfrac{3}{10}\)  |  (iii) \(\tfrac{7}{10}\)  |  (iv) \(\tfrac{4}{5}\)
Q2

Ten coins tossed simultaneously. Find \(P(\text{exactly 6 heads})\), \(P(\text{at least 6 heads})\), \(P(\text{at most 6 heads})\).

\(n=10, p=q=\tfrac{1}{2}\). Total outcomes \(=2^{10}=1024\)
(i) \(P(X=6)=\binom{10}{6}\!\left(\tfrac{1}{2}\right)^{10}=\dfrac{210}{1024}=\dfrac{105}{512}\)
(ii) \(P(X\ge6)=\dfrac{210+120+45+10+1}{1024}=\dfrac{386}{1024}=\dfrac{193}{512}\)
(iii) \(P(X\le6)=1-P(X\ge7)=1-\dfrac{176}{1024}=\dfrac{848}{1024}=\dfrac{53}{64}\)
(i) \(\dfrac{105}{512}\)  |  (ii) \(\dfrac{193}{512}\)  |  (iii) \(\dfrac{53}{64}\)
Q3CBSE 2023

Binomial \((n=5)\): \(P(X=1)=0.4096\) and \(P(X=2)=0.2048\). Find \(p\).

\(\dfrac{P(X=2)}{P(X=1)}=\dfrac{\binom{5}{2}p^2q^3}{\binom{5}{1}pq^4}=\dfrac{10p}{5q}=\dfrac{2p}{q}\)
\(\dfrac{0.2048}{0.4096}=0.5=\dfrac{2p}{q} \Rightarrow 2p=0.5q=0.5(1-p)\)
\(2p=0.5-0.5p \Rightarrow 2.5p=0.5 \Rightarrow p=0.2\)
\(p=0.2\) (i.e. \(\tfrac{1}{5}\))
Q4

Defective chips ~ Poisson(mean = 3). Find: (i) \(P(\text{exactly 2})\), (ii) \(P(\text{at most 2})\), (iii) \(P(\text{more than 2})\).

\(\lambda=3\)
(i) \(P(X=2)=\dfrac{e^{-3}\cdot9}{2}=\dfrac{9e^{-3}}{2}\)
(ii) \(P(X\le2)=e^{-3}\!\left(1+3+\tfrac{9}{2}\right)=\dfrac{17e^{-3}}{2}\)
(iii) \(P(X>2)=1-\dfrac{17e^{-3}}{2}\)
(i) \(\dfrac{9e^{-3}}{2}\)  |  (ii) \(\dfrac{17e^{-3}}{2}\)  |  (iii) \(1-\dfrac{17e^{-3}}{2}\)
Q5

800 students, height ~ \(N(165, 100)\). How many are (i) 155–175 cm, (ii) above 180 cm? [\(P(0<Z<1)=0.3413,\; P(0<Z<1.5)=0.4332\)]

\(\mu=165,\;\sigma=\sqrt{100}=10\)
(i) \(Z_1=\tfrac{155-165}{10}=-1,\; Z_2=\tfrac{175-165}{10}=1\). \(P(-1<Z<1)=0.6826\). Count \(=800\times0.6826\approx546\)
(ii) \(Z=\tfrac{180-165}{10}=1.5\). \(P(Z>1.5)=0.5-0.4332=0.0668\). Count \(=800\times0.0668\approx53\)
(i) ≈ 546 students  |  (ii) ≈ 53 students

Case Studies — Real-World Application Questions

4-mark case-based questions as per latest CBSE Class 12 Applied Maths pattern

Case Study 1Applied Maths

Photocopier Machines

An office has four copying machines. \(X\) = number in use at a given moment.
\(P(X=0)=0.10,\; P(X=1)=0.20,\; P(X=2)=0.30,\; P(X=3)=0.25,\; P(X=4)=0.15\)

(i) What is \(P(X\le2)\)?

  • (a) 0.50
  • (b) 0.60
  • (c) 0.70
  • (d) 0.80

✓ (b) 0.60

\(P(X\le2)=0.10+0.20+0.30=0.60\)
0.60

(ii) What is \(P(X>1)\)?

  • (a) 0.60
  • (b) 0.65
  • (c) 0.70
  • (d) 0.75

✓ (c) 0.70

\(P(X>1)=0.30+0.25+0.15=0.70\)
0.70

(iii)(a) Expected number of machines in use?

  • (a) 2.05
  • (b) 2.15
  • (c) 2.25
  • (d) 2.35

✓ (b) 2.15

\(E(X)=0(0.10)+1(0.20)+2(0.30)+3(0.25)+4(0.15)=0+0.20+0.60+0.75+0.60=2.15\)
\(E(X)=2.15\)

(iii)(b) Calculate variance and standard deviation.

✓ Var = 1.5275, SD ≈ 1.236

\(E(X^2)=0+0.20+1.20+2.25+2.40=6.05\)
\(\text{Var}(X)=6.05-(2.15)^2=6.05-4.6225=1.5275\)
\(\text{SD}=\sqrt{1.5275}\approx1.236\)
Var = 1.5275  |  SD ≈ 1.236
Case Study 2Applied Maths

Student Scores — Normal Distribution

Maths scores of 400 students ~ \(N(\mu=70, \sigma=10)\).

(i) Percentage who scored below 70 marks?

  • (a) 25%
  • (b) 40%
  • (c) 50%
  • (d) 60%

✓ (c) 50%

Normal distribution is symmetric about its mean — exactly 50% lie below \(\mu=70\).
50%

(ii) Students scoring above 80 marks? [\(P(0<Z<1)=0.3413\)]

  • (a) 58
  • (b) 63
  • (c) 68
  • (d) 73

✓ (b) 63

\(Z=\tfrac{80-70}{10}=1\). \(P(Z>1)=0.5-0.3413=0.1587\). Count \(=400\times0.1587\approx63\)
≈ 63 students

(iii)(a) Students scoring 60–80? [\(P(0<Z<1)=0.3413\)]

  • (a) 250
  • (b) 265
  • (c) 273
  • (d) 285

✓ (c) 273

\(P(-1<Z<1)=2\times0.3413=0.6826\). Count \(=400\times0.6826\approx273\)
≈ 273 students

(iii)(b) Top 5% receive certificates. Minimum qualifying score? [\(Z=1.645\)]

✓ Minimum score = 86.45

\(1.645=\dfrac{X-70}{10} \Rightarrow X=70+16.45=86.45\)
Minimum qualifying score = 86.45 marks
Case Study 3Applied Maths

Phone Calls at Reservation Desk

Calls arrive at 48 per hour at an airline desk — Poisson distribution.

(i) \(P(\text{exactly 3 calls in 5 minutes})\)?

  • (a) \(\tfrac{4e^{-4}}{3}\)
  • (b) \(\tfrac{16e^{-4}}{3}\)
  • (c) \(\tfrac{32e^{-4}}{3}\)
  • (d) \(\tfrac{64e^{-4}}{6}\)

✓ (c) \(\dfrac{32e^{-4}}{3}\)

\(\lambda_{\text{per min}}=\tfrac{48}{60}=0.8\). For 5 min: \(\lambda=5\times0.8=4\)
\(P(X=3)=\dfrac{e^{-4}\cdot4^3}{3!}=\dfrac{64e^{-4}}{6}=\dfrac{32e^{-4}}{3}\)
\(P(X=3)=\dfrac{32e^{-4}}{3}\)

(ii) Agent takes 5 min per call. Expected calls waiting? \(P(\text{none waiting})\)?

✓ Expected = 4  |  \(P(\text{none})=e^{-4}\)

\(\lambda=4\) for 5 minutes. \(E(X)=\lambda=4\). \(P(X=0)=e^{-4}\)
Expected = 4 calls  |  \(P(\text{none})=e^{-4}\)

(iii) \(P(\text{agent takes 3-min break without interruption})\)?

  • (a) \(e^{-2.4}\)
  • (b) \(e^{-3}\)
  • (c) \(e^{-3.6}\)
  • (d) \(e^{-4}\)

✓ (a) \(e^{-2.4}\)

\(\lambda_{\text{3 min}}=3\times0.8=2.4\). \(P(X=0)=e^{-2.4}\)
\(P=e^{-2.4}\)

Exam Tips for Unit 4 — Probability Distributions

Common mistakes examiners flag every year

📊

Always verify \(\sum P(X=x)=1\)

Before computing any mean or variance, check that all probabilities sum to exactly 1. If a \(k\) is unknown, this is always your first equation. Examiners deduct marks if this verification step is missing.

Frequently Asked Questions — Unit 4 Probability Distributions

Common questions students ask about Class 12 Applied Maths Probability Distributions

Unit 4 – Probability Distributions carries 10 marks in the CBSE Class 12 Applied Mathematics board exam. Questions typically appear as 1-mark MCQs, 2–3 mark short answers, and 4-mark case studies covering all four sub-topics.

Unit 4 covers four key topics: (1) Random Variable and Probability Distribution Table — discrete and continuous variables, E(X) and Var(X); (2) Binomial Distribution — Bernoulli trials, PMF \(P(X=x)=\binom{n}{x}p^xq^{n-x}\), mean \(=np\), variance \(=npq\); (3) Poisson Distribution — PMF \(P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}\), mean = variance = λ; and (4) Normal Distribution — z-score \(Z=\frac{X-\mu}{\sigma}\), area under the standard normal curve.

Use \(\text{Var}(X) = E(X^2) - [E(X)]^2\). First calculate \(E(X) = \sum x \cdot P(X=x)\), then \(E(X^2) = \sum x^2 \cdot P(X=x)\), then subtract \([E(X)]^2\). Example: if \(P(X=0)=0.3, P(X=1)=0.4, P(X=2)=0.3\), then \(E(X)=0+0.4+0.6=1\), \(E(X^2)=0+0.4+1.2=1.6\), \(\text{Var}(X)=1.6-1=0.6\).

For Binomial Distribution \(X \sim B(n, p)\): Mean \(= np\) and Variance \(= npq\), where \(q = 1-p\). The standard deviation is \(\sigma = \sqrt{npq}\). Example: \(B(10, 0.3)\) has mean \(= 10 \times 0.3 = 3\) and variance \(= 10 \times 0.3 \times 0.7 = 2.1\).

\(P(X=x) = \dfrac{e^{-\lambda}\lambda^x}{x!}\) where \(\lambda\) is both the mean and the variance. Poisson distribution is used when: n is very large, p is very small, and np = λ is finite. Common applications: phone calls per minute, defects per batch, accidents per day. For Poisson, \(\text{SD} = \sqrt{\lambda}\).

\(Z = \dfrac{X - \mu}{\sigma}\) converts any normal variable \(X \sim N(\mu, \sigma^2)\) to the standard normal \(Z \sim N(0,1)\). Example: if \(X \sim N(70, 100)\) and you want \(P(X > 80)\): \(Z = \frac{80-70}{10} = 1\), so \(P(X>80) = P(Z>1) = 0.5 - P(0<Z<1) = 0.5 - 0.3413 = 0.1587\).

The most frequent mistake is applying \(\text{Var}(aX+b) = a^2\text{Var}(X) + b\) instead of the correct \(\text{Var}(aX+b) = a^2\text{Var}(X)\) — the constant \(b\) has no effect on variance. Students also commonly confuse \(\sigma^2 = npq\) (variance) with \(\sigma = npq\) — remember \(\sigma\) is the square root of npq. Another error: forgetting to verify \(\sum P(X=x) = 1\) when finding \(k\).

Based on recent papers, the most frequently asked topics are: Finding k in probability distributions (appears every year as MCQ or short answer — always set \(\sum P(X=x)=1\)), Mean and variance of binomial distribution (MCQ level), Poisson probability calculations (short answer), and z-score and normal distribution area problems (long answer or case study). Make sure to memorise the Poisson and binomial formulas exactly as they appear.

Explore All Units — Class 12 Applied Maths

Free study material for all 8 units of the CBSE Applied Maths syllabus

Unit 1Numbers & Quantification11 Marks Unit 2Algebra10 Marks Unit 3Calculus15 Marks Unit 4 · CurrentProbability Distributions10 Marks Unit 5Inferential Statistics5 Marks Unit 6Index Numbers & Time Based Data6 Marks Unit 7Financial Mathematics10 Marks Unit 8Linear Programming8 Marks All UnitsCourse Overview80 Marks