How Probability Appears in the Board Exam

Probability is among the most predictable chapters in the CBSE paper. The typical spread:

The 5-mark Bayes question is the most reliable long question in the entire paper. The marking scheme rewards the template: define events (1 mark), write priors and conditionals (1 to 2 marks), state the theorem (1 mark), substitute and simplify (1 to 2 marks). Write every step even when the arithmetic feels easy.

Below are 12 board-style written questions with complete solutions, organised by marks, followed by a 15-question MCQ quiz.

2-Mark Questions

Question 1 — Conditionals from partial data

Given $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find (i) $P(A \cap B)$, (ii) $P(A|B)$, (iii) $P(A \cup B)$.

Step 1 — multiplication theorem: $P(A \cap B) = P(A)\,P(B|A) = 0.8 \times 0.4 = 0.32$.

Step 2 — reverse conditional: $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.32}{0.5} = 0.64$.

Step 3 — addition rule: $P(A \cup B) = 0.8 + 0.5 - 0.32 = 0.98$.

Answer: (i) $0.32$, (ii) $0.64$, (iii) $0.98$.

Question 2 — Complements of independent events

If $A$ and $B$ are independent events, prove that $A'$ and $B'$ are also independent.

Step 1 — De Morgan plus the complement rule: $$P(A' \cap B') = P\big((A \cup B)'\big) = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)$$

Step 2 — use independence $P(A \cap B) = P(A)P(B)$: $$P(A' \cap B') = 1 - P(A) - P(B) + P(A)\,P(B)$$

Step 3 — factor: $$= \big(1 - P(A)\big)\big(1 - P(B)\big) = P(A')\,P(B')$$

which is the independence condition for $A'$ and $B'$. $\blacksquare$

Question 3 — Conditioning on earlier throws

A die is thrown three times. $E$ is the event "4 appears on the third throw" and $F$ is the event "6 appears on the first throw and 5 on the second". Find $P(E|F)$.

Step 1 — reduced sample space: given $F$, the first two throws are fixed at $(6, 5)$ and only the third throw varies over $\{1, 2, 3, 4, 5, 6\}$.

Step 2 — favourable: the third throw is 4 in exactly one of those six outcomes.

Answer: $P(E|F) = \dfrac{1}{6}$ — the third throw is unaffected by the first two.

Question 4 — Independence in one draw

A card is drawn from a well-shuffled deck of 52. $E$ is the event "the card is a king" and $F$ the event "the card is red". Check whether $E$ and $F$ are independent.

Step 1 — the three numbers: $$P(E) = \frac{4}{52} = \frac{1}{13}, \qquad P(F) = \frac{26}{52} = \frac{1}{2}, \qquad P(E \cap F) = P(\text{red king}) = \frac{2}{52} = \frac{1}{26}$$

Step 2 — product test: $P(E)\,P(F) = \frac{1}{13} \times \frac{1}{2} = \frac{1}{26} = P(E \cap F)$.

Answer: the events are independent — kings are split evenly between the colours.

3-Mark Questions

Question 5 — When do they contradict each other?

$A$ speaks the truth in $60\%$ of cases and $B$ in $90\%$ of cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

Step 1 — contradiction happens in two disjoint ways: $A$ truthful and $B$ lying, or $A$ lying and $B$ truthful. The speakers are independent.

Step 2 — compute each way: $$P(\text{contradict}) = P(A_T)\,P(B_L) + P(A_L)\,P(B_T) = 0.6 \times 0.1 + 0.4 \times 0.9 = 0.06 + 0.36 = 0.42$$

Answer: they contradict each other in $42\%$ of cases.

Question 6 — Three spades in a row

Three cards are drawn successively without replacement from a deck of 52. Find the probability that all three are spades.

Step 1 — three-event multiplication theorem: $$P(S_1 \cap S_2 \cap S_3) = P(S_1)\,P(S_2|S_1)\,P(S_3|S_1 \cap S_2)$$

Step 2 — substitute: $$= \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} = \frac{1}{4} \times \frac{4}{17} \times \frac{11}{50} = \frac{11}{850}$$

Answer: $\dfrac{11}{850}$.

Question 7 — Solving for an unknown probability

$A$ and $B$ are independent events with $P(A) = 0.2$ and $P(A \cup B) = 0.6$. Find $P(B)$.

Step 1 — union with the independence substitution: with $P(B) = p$, $$P(A \cup B) = P(A) + p - P(A)\,p \implies 0.6 = 0.2 + p - 0.2p$$

Step 2 — solve: $0.4 = 0.8p$, so $p = 0.5$.

Answer: $P(B) = 0.5$. (Verify: $0.2 + 0.5 - 0.1 = 0.6$ ✓)

Question 8 — Total probability over two bags

Bag I contains 4 red and 4 black balls; Bag II contains 2 red and 6 black balls. A bag is selected at random and one ball is drawn from it. Find the probability that the ball is red.

Step 1 — partition by the bag: $P(E_1) = P(E_2) = \frac{1}{2}$.

Step 2 — conditionals: $P(R|E_1) = \frac{4}{8} = \frac{1}{2}$, $P(R|E_2) = \frac{2}{8} = \frac{1}{4}$.

Step 3 — total probability: $$P(R) = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$$

Answer: $\dfrac{3}{8}$.

5-Mark Questions (Bayes' Theorem)

Question 9 — Urban or rural?

In a district, $60\%$ of families are urban and $40\%$ are rural. $20\%$ of urban families and $10\%$ of rural families own a two-wheeler. A family chosen at random owns a two-wheeler. Find the probability that it is an urban family.

Step 1 — define events: $U$ = urban, $R$ = rural, $T$ = owns a two-wheeler. $\{U, R\}$ partitions the population.

Step 2 — priors and conditionals: $P(U) = 0.6$, $P(R) = 0.4$; $P(T|U) = 0.2$, $P(T|R) = 0.1$.

Step 3 — Bayes' theorem: $$P(U|T) = \frac{P(U)\,P(T|U)}{P(U)\,P(T|U) + P(R)\,P(T|R)} = \frac{0.6 \times 0.2}{0.6 \times 0.2 + 0.4 \times 0.1} = \frac{0.12}{0.16} = \frac{3}{4}$$

Answer: the probability that the two-wheeler-owning family is urban is $\dfrac{3}{4}$.

Question 10 — Which machine produced the defective item?

Machines A, B and C produce $30\%$, $50\%$ and $20\%$ of a factory's items, with $1\%$, $2\%$ and $3\%$ of their outputs defective. An item drawn at random is defective. Find the probability that it was produced by machine C.

Step 1 — define events: $E_1, E_2, E_3$ = item from A, B, C; $D$ = defective. The $E_i$ partition the output.

Step 2 — priors and conditionals: $P(E_1) = 0.3$, $P(E_2) = 0.5$, $P(E_3) = 0.2$; $P(D|E_1) = 0.01$, $P(D|E_2) = 0.02$, $P(D|E_3) = 0.03$.

Step 3 — Bayes' theorem: $$P(E_3|D) = \frac{0.2 \times 0.03}{0.3 \times 0.01 + 0.5 \times 0.02 + 0.2 \times 0.03} = \frac{0.006}{0.003 + 0.010 + 0.006} = \frac{6}{19}$$

Answer: $\dfrac{6}{19}$. Machine C makes only a fifth of the items but has the worst defect rate, so its posterior share of defectives ($\approx 32\%$) exceeds its production share.

Question 11 — The truthful reporter

A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Step 1 — hypotheses about the die: $S_1$ = six occurred ($P = \frac{1}{6}$), $S_2$ = six did not occur ($P = \frac{5}{6}$).

Step 2 — conditionals for $E$ = "he reports a six": $P(E|S_1) = \frac{4}{5}$ (truth), $P(E|S_2) = \frac{1}{5}$ (lie).

Step 3 — Bayes' theorem: $$P(S_1|E) = \frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}} = \frac{\frac{4}{30}}{\frac{4}{30} + \frac{5}{30}} = \frac{4}{9}$$

Answer: $\dfrac{4}{9}$. Even at $80\%$ honesty the report of a six is more likely false than true, because a six is rare while a lie about a non-six is common.

Question 12 — Reasoning backwards through a two-stage experiment

A girl throws a die. If the outcome is 5 or 6, she tosses a coin three times and notes the number of heads; if it is 1, 2, 3 or 4, she tosses the coin once. Given that she obtained exactly one head, find the probability that the die showed 5 or 6.

Step 1 — hypotheses from the die: $E_1$ = "5 or 6" ($P = \frac{1}{3}$), $E_2$ = "1, 2, 3 or 4" ($P = \frac{2}{3}$).

Step 2 — conditionals for $H$ = "exactly one head": $P(H|E_1) = \frac{3}{8}$ (one head in three tosses), $P(H|E_2) = \frac{1}{2}$ (one toss).

Step 3 — Bayes' theorem: $$P(E_1|H) = \frac{\frac{1}{3} \times \frac{3}{8}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{8}}{\frac{11}{24}} = \frac{3}{11}$$

Answer: $\dfrac{3}{11}$.

Presentation tip for all four questions above: every solution follows the same five-line skeleton — events, priors, conditionals, theorem, answer-in-words. Board examiners award partial credit line by line; the skeleton guarantees you collect it.