CBSE Class 12 · Chapter 13 · Probability · 2026-27

Probability Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceConditional probability

If $P (A \cap B) = 0.2$ and $P (B) = 0.5$ , then $P (A ∣ B)$ is

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Answer: (a)

0.4

Why: P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5.

$P (A ∣ B) = \frac{0.2}{0.5} = 0.4$ .

Q2

·1 mark·Multiple choiceMultiplication theorem and independent events

Two independent events have probabilities $P (A) = 0.3$ and $P (B) = 0.5$ . What is the probability that both occur?

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Answer: (b)

0.15

Why: Independent events: P(A ∩ B) = P(A)P(B).

$P (A \cap B) = 0.3 \times 0.5 = 0.15$ .

Q3

·1 mark·Multiple choiceMultiplication theorem and independent events

$A$ and $B$ are independent with $P (A) = \frac{1}{2}$ and $P (B) = \frac{1}{3}$ . Then $P (A \cup B)$ is

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Answer: (c)

\frac{2}{3}

Why: P(A ∪ B) = P(A) + P(B) − P(A)P(B).

$P (A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{2} \cdot \frac{1}{3} = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}$ .

Q4

·1 mark·Multiple choiceMultiplication theorem and independent events

Two cards are drawn one after the other, without replacement, from a well-shuffled pack of $52$ . The probability that the first is a king and the second is a queen is

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Answer: (b)

\frac{4}{663}

Why: P = (4/52) × (4/51).

$P (K_{1}) \times P (Q_{2} ∣ K_{1}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}$ . ( $\frac{1}{169}$ would be with replacement; $\frac{8}{663}$ counts either order.)

Q5

·1 mark·Multiple choiceConditional probability

A die is thrown. Given that the number shown is even, the probability that it is greater than $3$ is

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Answer: (d)

\frac{2}{3}

Why: Among 2, 4, 6, two numbers (4, 6) exceed 3.

$B = {2, 4, 6}$ , $A \cap B = {4, 6}$ : $P (A ∣ B) = \frac{2 / 6}{3 / 6} = \frac{2}{3}$ .

Q6

·1 mark·Multiple choiceTotal probability and Bayes' theorem

Bag I has $3$ red and $2$ black balls; bag II has $1$ red and $4$ black balls. A bag is chosen at random and a ball drawn from it is red. The probability that it came from bag I is

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Answer: (a)

\frac{3}{4}

Why: Bayes: (½ × 3/5) / (½ × 3/5 + ½ × 1/5).

$P (I ∣ R) = \frac{\frac{1}{2} \cdot \frac{3}{5}}{\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{1}{5}} = \frac{3 / 10}{4 / 10} = \frac{3}{4}$ .

Q7

·1 mark·Multiple choiceTotal probability and Bayes' theorem

With the same two bags (bag I: $3$ red, $2$ black; bag II: $1$ red, $4$ black) and a bag chosen at random, the probability of drawing a red ball is

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Answer: (c)

\frac{2}{5}

Why: Total probability: ½ × 3/5 + ½ × 1/5.

$P (R) = \frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{1}{5} = \frac{3}{10} + \frac{1}{10} = \frac{2}{5}$ .

Q8

·1 mark·Multiple choiceRandom variable and its mean

A random variable $X$ takes the values $1$ , $2$ and $3$ with probabilities $0.2$ , $0.5$ and $0.3$ . The mean of $X$ is

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Answer: (d)

2.1

Why: E(X) = Σ x p = 0.2 + 1.0 + 0.9.

$E (X) = 1 (0.2) + 2 (0.5) + 3 (0.3) = 0.2 + 1 + 0.9 = 2.1$ .

Q9

·1 mark·Multiple choiceRandom variable and its mean

A random variable $X$ has $P (X = x) = k x$ for $x = 1, 2, 3, 4$ (and $0$ otherwise). The value of $k$ is

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Answer: (b)

\frac{1}{10}

Why: Probabilities add to 1: k(1 + 2 + 3 + 4) = 1.

$k + 2 k + 3 k + 4 k = 10 k = 1 \Rightarrow k = \frac{1}{10}$ .

Q10

·1 mark·Multiple choiceMultiplication theorem and independent events

If $A$ and $B$ are independent events, then $A$ and $B^{'}$ are

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Answer: (c) independent

Why: P(A ∩ B′) = P(A) − P(A)P(B) = P(A)P(B′).

$P (A \cap B^{'}) = P (A) - P (A \cap B) = P (A) - P (A) P (B) = P (A) (1 - P (B)) = P (A) P (B^{'})$ , so $A$ and $B^{'}$ are independent.

Q11

·1 mark·Multiple choiceMultiplication theorem and independent events

If $P (A) = 0.6$ and $P (B ∣ A) = 0.25$ , then $P (A \cap B)$ is

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Answer: (a)

0.15

Why: Multiplication theorem: P(A ∩ B) = P(A) P(B|A).

$P (A \cap B) = P (A) P (B ∣ A) = 0.6 \times 0.25 = 0.15$ .

Q12

·1 mark·Multiple choiceConditional probability

Three fair coins are tossed. Given that at least two of them show heads, the probability that all three show heads is

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Answer: (b)

\frac{1}{4}

Why: At least two heads: HHT, HTH, THH, HHH; only HHH has three.

$B = {H H T, H T H, T H H, H H H}$ ( $4$ of the $8$ outcomes), $A \cap B = {H H H}$ . $P (A ∣ B) = \frac{1 / 8}{4 / 8} = \frac{1}{4}$ .

Q13

·1 mark·Multiple choiceConditional probability

$A$ and $B$ are mutually exclusive events with $P (A) > 0$ and $P (B) > 0$ . Then $P (A ∣ B)$ equals

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Answer: (c)

0

Why: Mutually exclusive: A ∩ B = ∅, so P(A ∩ B) = 0.

$P (A ∣ B) = \frac{P (A \cap B)}{P (B)} = \frac{0}{P (B)} = 0$ .

Q14

·1 mark·Multiple choiceRandom variable and its mean

A fair coin is tossed twice and $X$ is the number of heads. The mean of $X$ is

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Answer: (b)

1

Why: P(X = 0, 1, 2) = 1/4, 1/2, 1/4; E(X) = 0 + 1/2 + 2/4.

$E (X) = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} = 1$ .

Q15

·1 mark·Multiple choiceMultiplication theorem and independent events

Three archers, independently, hit a target with probabilities $\frac{1}{2}$ , $\frac{1}{3}$ and $\frac{1}{4}$ . The probability that the target is hit at least once is

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Answer: (a)

\frac{3}{4}

Why: 1 − P(all miss) = 1 − (1/2)(2/3)(3/4).

$P (all miss) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$ , so $P (at least one hit) = 1 - \frac{1}{4} = \frac{3}{4}$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Two filling machines (4 marks)

A bottling plant fills juice bottles on two machines. Machine A fills $60 %$ of the bottles and machine B the rest. Quality checks show that $2 %$ of the bottles filled by A and $7 %$ of those filled by B are under-filled. A bottle is picked at random from a day's output.

(i) Find the probability that the bottle was filled by B and is under-filled. [1 mark]

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Answer:

0.028

P (B \cap U) = 0.4 \times 0.07 = 0.028

. A1

(ii) Find the probability that the bottle is under-filled. [1 mark]

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Answer:

0.04

Total probability:

P (U) = 0.6 (0.02) + 0.4 (0.07) = 0.012 + 0.028 = 0.04

. A1

(iii) An under-filled bottle is found. Find the probability that machine B filled it. [2 marks]

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Answer:

\frac{7}{10}

Bayes:

P (B ∣ U) = \frac{0.028}{0.04}

= \frac{7}{10}

. A1

OR Machine B cannot be improved. To what percentage must machine A's under-fill rate be reduced for the plant's overall under-fill rate to be

3.4 %

? [2 marks]

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Answer:

1 %

0.6 p + 0.028 = 0.034

\Rightarrow 0.6 p = 0.006 \Rightarrow p = 0.01

, i.e.

1 %

. A1

Case study 2: Choosing a quiz team (4 marks)

For an inter-school quiz, two students are chosen at random from a group of $4$ boys and $2$ girls. Let $X$ be the number of girls chosen.

(i) Find

P (X = 2)

. [1 mark]

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Answer:

\frac{1}{15}

P (X = 2) = \frac{^{2} C_{2}}{^{6} C_{2}} = \frac{1}{15}

. A1

(ii) Find

P (X = 0)

. [1 mark]

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Answer:

\frac{2}{5}

P (X = 0) = \frac{^{4} C_{2}}{^{6} C_{2}} = \frac{6}{15} = \frac{2}{5}

. A1

(iii) Write the probability distribution of

X

and find its mean. [2 marks]

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Answer:

\frac{2}{5}, \frac{8}{15}, \frac{1}{15}

; mean

\frac{2}{3}

P (X = 1) = \frac{4 \times 2}{15} = \frac{8}{15}

, so

X = 0, 1, 2

with

\frac{6}{15}, \frac{8}{15}, \frac{1}{15}

M1;

E (X) = \frac{8}{15} + \frac{2}{15} = \frac{2}{3}

. A1

OR Find the probability that at least one girl is chosen, and the probability that exactly one girl is chosen given that at least one girl is chosen. [2 marks]

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Answer:

\frac{3}{5}

;

\frac{8}{9}

P (X \geq 1) = 1 - \frac{2}{5} = \frac{3}{5}

M1;

P (X = 1 ∣ X \geq 1) = \frac{8 / 15}{9 / 15} = \frac{8}{9}

. A1

Next steps for Probability

This free set is separate from the chapter's question bank. On the Probability chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 56 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Probability: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.