If \(P(A \cap B) = 0.2\) and \(P(B) = 0.5\), then \(P(A \mid B)\) is
- (a)\(0.4\)
- (b)\(0.1\)
- (c)\(0.25\)
- (d)\(0.7\)
Show answer
Answer: (a) \(0.4\)
Why: P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5.
\(P(A \mid B) = \dfrac{0.2}{0.5} = 0.4\).
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)
- No calculator needed
- Free, no sign-in
Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
Two independent events have probabilities \(P(A) = 0.3\) and \(P(B) = 0.5\). What is the probability that both occur?
- (a)\(0.8\)
- (b)\(0.15\)
- (c)\(0.2\)
- (d)\(0.35\)
Show answer
Answer: (b) \(0.15\)
Why: Independent events: P(A ∩ B) = P(A)P(B).
\(P(A \cap B) = 0.3 \times 0.5 = 0.15\).
\(A\) and \(B\) are independent with \(P(A) = \dfrac12\) and \(P(B) = \dfrac13\). Then \(P(A \cup B)\) is
- (a)\(\dfrac56\)
- (b)\(\dfrac16\)
- (c)\(\dfrac23\)
- (d)\(\dfrac12\)
Show answer
Answer: (c) \(\dfrac23\)
Why: P(A ∪ B) = P(A) + P(B) − P(A)P(B).
\(P(A \cup B) = \dfrac12 + \dfrac13 - \dfrac12 \cdot \dfrac13 = \dfrac56 - \dfrac16 = \dfrac23\).
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
- (a)\(\dfrac{1}{169}\)
- (b)\(\dfrac{4}{663}\)
- (c)\(\dfrac{1}{221}\)
- (d)\(\dfrac{8}{663}\)
Show answer
Answer: (b) \(\dfrac{4}{663}\)
Why: P = (4/52) × (4/51).
\(P(K_1) \times P(Q_2 \mid K_1) = \dfrac{4}{52} \times \dfrac{4}{51} = \dfrac{16}{2652} = \dfrac{4}{663}\). (\(\tfrac{1}{169}\) would be with replacement; \(\tfrac{8}{663}\) counts either order.)
A die is thrown. Given that the number shown is even, the probability that it is greater than \(3\) is
- (a)\(\dfrac12\)
- (b)\(\dfrac13\)
- (c)\(\dfrac34\)
- (d)\(\dfrac23\)
Show answer
Answer: (d) \(\dfrac23\)
Why: Among 2, 4, 6, two numbers (4, 6) exceed 3.
\(B = \{2, 4, 6\}\), \(A \cap B = \{4, 6\}\): \(P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac23\).
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
- (a)\(\dfrac34\)
- (b)\(\dfrac14\)
- (c)\(\dfrac35\)
- (d)\(\dfrac12\)
Show answer
Answer: (a) \(\dfrac34\)
Why: Bayes: (½ × 3/5) / (½ × 3/5 + ½ × 1/5).
\(P(\text{I} \mid R) = \dfrac{\frac12 \cdot \frac35}{\frac12 \cdot \frac35 + \frac12 \cdot \frac15} = \dfrac{3/10}{4/10} = \dfrac34\).
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
- (a)\(\dfrac{3}{10}\)
- (b)\(\dfrac12\)
- (c)\(\dfrac25\)
- (d)\(\dfrac45\)
Show answer
Answer: (c) \(\dfrac25\)
Why: Total probability: ½ × 3/5 + ½ × 1/5.
\(P(R) = \dfrac12 \cdot \dfrac35 + \dfrac12 \cdot \dfrac15 = \dfrac{3}{10} + \dfrac{1}{10} = \dfrac25\).
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
- (a)\(2\)
- (b)\(1.9\)
- (c)\(2.5\)
- (d)\(2.1\)
Show answer
Answer: (d) \(2.1\)
Why: E(X) = Σ x p = 0.2 + 1.0 + 0.9.
\(E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1 + 0.9 = 2.1\).
A random variable \(X\) has \(P(X = x) = kx\) for \(x = 1, 2, 3, 4\) (and \(0\) otherwise). The value of \(k\) is
- (a)\(\dfrac14\)
- (b)\(\dfrac{1}{10}\)
- (c)\(\dfrac15\)
- (d)\(\dfrac16\)
Show answer
Answer: (b) \(\dfrac{1}{10}\)
Why: Probabilities add to 1: k(1 + 2 + 3 + 4) = 1.
\(k + 2k + 3k + 4k = 10k = 1 \Rightarrow k = \dfrac{1}{10}\).
If \(A\) and \(B\) are independent events, then \(A\) and \(B'\) are
- (a)mutually exclusive
- (b)always dependent
- (c)independent
- (d)exhaustive
Show answer
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.
If \(P(A) = 0.6\) and \(P(B \mid A) = 0.25\), then \(P(A \cap B)\) is
- (a)\(0.15\)
- (b)\(0.85\)
- (c)\(0.35\)
- (d)\(0.24\)
Show answer
Answer: (a) \(0.15\)
Why: Multiplication theorem: P(A ∩ B) = P(A) P(B|A).
\(P(A \cap B) = P(A)\,P(B \mid A) = 0.6 \times 0.25 = 0.15\).
Three fair coins are tossed. Given that at least two of them show heads, the probability that all three show heads is
- (a)\(\dfrac12\)
- (b)\(\dfrac14\)
- (c)\(\dfrac18\)
- (d)\(\dfrac13\)
Show answer
Answer: (b) \(\dfrac14\)
Why: At least two heads: HHT, HTH, THH, HHH; only HHH has three.
\(B = \{HHT, HTH, THH, HHH\}\) (\(4\) of the \(8\) outcomes), \(A \cap B = \{HHH\}\). \(P(A \mid B) = \dfrac{1/8}{4/8} = \dfrac14\).
\(A\) and \(B\) are mutually exclusive events with \(P(A) \gt 0\) and \(P(B) \gt 0\). Then \(P(A \mid B)\) equals
- (a)\(1\)
- (b)\(P(A)\)
- (c)\(0\)
- (d)\(\dfrac{P(A)}{P(B)}\)
Show answer
Answer: (c) \(0\)
Why: Mutually exclusive: A ∩ B = ∅, so P(A ∩ B) = 0.
\(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0}{P(B)} = 0\).
A fair coin is tossed twice and \(X\) is the number of heads. The mean of \(X\) is
- (a)\(\dfrac12\)
- (b)\(1\)
- (c)\(2\)
- (d)\(\dfrac32\)
Show answer
Answer: (b) \(1\)
Why: P(X = 0, 1, 2) = 1/4, 1/2, 1/4; E(X) = 0 + 1/2 + 2/4.
\(E(X) = 0 \cdot \dfrac14 + 1 \cdot \dfrac12 + 2 \cdot \dfrac14 = 1\).
Three archers, independently, hit a target with probabilities \(\dfrac12\), \(\dfrac13\) and \(\dfrac14\). The probability that the target is hit at least once is
- (a)\(\dfrac34\)
- (b)\(\dfrac{1}{24}\)
- (c)\(\dfrac14\)
- (d)\(\dfrac{23}{24}\)
Show answer
Answer: (a) \(\dfrac34\)
Why: 1 − P(all miss) = 1 − (1/2)(2/3)(3/4).
\(P(\text{all miss}) = \dfrac12 \cdot \dfrac23 \cdot \dfrac34 = \dfrac14\), so \(P(\text{at least one hit}) = 1 - \dfrac14 = \dfrac34\).
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.
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]
Show answer
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]
Show answer
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]
Show answer
Answer: \(\dfrac{7}{10}\)
Bayes: \(P(B \mid U) = \dfrac{0.028}{0.04}\) M1 \(= \dfrac{7}{10}\). 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]
Show answer
Answer: \(1\%\)
\(0.6p + 0.028 = 0.034\) M1 \(\Rightarrow 0.6p = 0.006 \Rightarrow p = 0.01\), 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]
Show answer
Answer: \(\dfrac{1}{15}\)
\(P(X = 2) = \dfrac{^2C_2}{^6C_2} = \dfrac{1}{15}\). A1
(ii) Find \(P(X = 0)\). [1 mark]
Show answer
Answer: \(\dfrac25\)
\(P(X = 0) = \dfrac{^4C_2}{^6C_2} = \dfrac{6}{15} = \dfrac25\). A1
(iii) Write the probability distribution of \(X\) and find its mean. [2 marks]
Show answer
Answer: \(\dfrac25, \dfrac{8}{15}, \dfrac{1}{15}\); mean \(\dfrac23\)
\(P(X = 1) = \dfrac{4 \times 2}{15} = \dfrac{8}{15}\), so \(X = 0, 1, 2\) with \(\dfrac{6}{15}, \dfrac{8}{15}, \dfrac{1}{15}\) M1; \(E(X) = \dfrac{8}{15} + \dfrac{2}{15} = \dfrac23\). 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]
Show answer
Answer: \(\dfrac35\); \(\dfrac89\)
\(P(X \ge 1) = 1 - \dfrac25 = \dfrac35\) M1; \(P(X = 1 \mid X \ge 1) = \dfrac{8/15}{9/15} = \dfrac89\). 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 Essentials adds all 39 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 and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
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 re-checked by computer algebra and 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.