Sample 2026-27 Class 12 Maths Q18 Solution: Find k = P(B) when exactly one of two independent events has probability 1/3

The question

$A$ and $B$ are two independent events such that $P(A) = \frac{1}{6}$ ⁠, $P(B) = k$ ⁠. If the probability of occurrence of exactly one of them is $\frac{1}{3}$ ⁠, then find the value of $k$ ⁠.

(A) $\frac{1}{4}$

(B) $\frac{2}{3}$

(C) $\frac{3}{4}$

(D) $\frac{5}{6}$

Exam answer

$A$ and $B$ are independent, so P(exactly one) is

\begin{aligned} &P(A' \cap B) + P(A \cap B') \\ &= P(A')P(B) + P(A)P(B') \\ &= \left(1 - \frac{1}{6}\right)k + \frac{1}{6}(1 - k) \end{aligned}

Why?

Exactly one happens in two ways: $B$ without $A$ ⁠, or $A$ without $B$ ⁠. These cannot both happen, so their probabilities add.

If $A$ and $B$ are independent, so are $A'$ and $B$ ⁠, and $A$ and $B'$ ⁠. That lets us multiply. Also $P(A') = 1 - P(A)$ and $P(B') = 1 - k$ ⁠.

1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}

\begin{aligned} &\frac{5}{6}k + \frac{1}{6}(1 - k) = \frac{1}{3} \\ \Rightarrow\ &\frac{5}{6}k + \frac{1}{6} - \frac{1}{6}k = \frac{1}{3} \\ \Rightarrow\ &k\left(\frac{5}{6} - \frac{1}{6}\right) = \frac{1}{3} - \frac{1}{6} \\ \Rightarrow\ &\frac{2}{3}k = \frac{1}{6} \Rightarrow k = \frac{1}{4} \end{aligned}

Why?

First open the bracket:

\frac{1}{6}(1 - k) = \frac{1}{6} - \frac{1}{6}k

Collect the $k$ terms on the left and the numbers on the right:

\begin{aligned} \frac{5}{6} - \frac{1}{6} &= \frac{4}{6} = \frac{2}{3} \\ \frac{1}{3} - \frac{1}{6} &= \frac{1}{6} \end{aligned}

In detail: $\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3}$ ⁠, and $\frac{1}{3} = \frac{2}{6}$ ⁠, so $\frac{2}{6} - \frac{1}{6} = \frac{1}{6}$ ⁠.

Then $k = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}$ ⁠. Check: $\frac{5}{24} + \frac{3}{24} = \frac{8}{24} = \frac{1}{3}$ ⁠.

\begin{aligned} \frac{2}{3}k &= \frac{1}{6} \\ k &= \frac{1}{6} \div \frac{2}{3} = \frac{1}{6} \times \frac{3}{2} \\ &= \frac{3}{12} = \frac{3 \div 3}{12 \div 3} = \frac{1}{4} \end{aligned}

The check, with $k = \frac{1}{4}$ and $1 - k = \frac{3}{4}$ ⁠:

\begin{aligned} \frac{5}{6} \times \frac{1}{4} &= \frac{5}{24} \\ \frac{1}{6} \times \frac{3}{4} &= \frac{3}{24} \\ \frac{5}{24} + \frac{3}{24} &= \frac{8}{24} = \frac{1}{3} \end{aligned}

Answer: (A) $\dfrac{1}{4}$

Where this comes from

Chapter 13, Probability: Independent events.

Question type: At least one, or exactly one

How to spot it: click to see

Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: The events are independent, and "exactly one" is $\frac13$ ⁠. Write it as $P(A')P(B) + P(A)P(B') = \frac56 k + \frac16(1 - k)$ ⁠, set it equal to $\frac13$ ⁠, and solve for $k$ ⁠.

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