Board 2024 Class 12 Maths Q11 Solution: If P(A|B) = P(A′|B), then P(A ∩ B) = ½ P(B)

The question

If $P(A|B) = P(A'|B)$ ⁠, then which of the following statements is true ?

(A) $P(A) = P(A')$

(B) $P(A) = 2\,P(B)$

(C) $P(A \cap B) = \frac{1}{2}\,P(B)$

(D) $P(A \cap B) = 2\,P(B)$

Exam answer

We know $P(A|B) + P(A'|B) = 1$ ⁠.

Why?

Once $B$ has happened, either $A$ happens or it does not. In terms of the definition,

\begin{aligned} &P(A|B) + P(A'|B) \\ &= \frac{P(A \cap B) + P(A' \cap B)}{P(B)} \\ &= \frac{P(B)}{P(B)} = 1 \end{aligned}

because $A \cap B$ and $A' \cap B$ split $B$ into two parts with no overlap.

Given $P(A|B) = P(A'|B)$ ⁠,

\begin{aligned} P(A|B) + P(A'|B) &= 1 \\ \Rightarrow 2P(A|B) &= 1 \\ \Rightarrow \frac{P(A \cap B)}{P(B)} &= \frac{1}{2} \\ \Rightarrow P(A \cap B) &= \frac{1}{2}P(B) \end{aligned}

Why?

The two equal terms add to $2P(A|B)$ ⁠. Divide both sides by 2:

\begin{aligned} P(A|B) + P(A|B) &= 1 \\ 2P(A|B) &= 1 \\ P(A|B) &= \frac12 \end{aligned}

Then the definition of conditional probability is used:

P(A|B) = \frac{P(A \cap B)}{P(B)}

Multiplying both sides by $P(B)$ gives the last line.

The condition says nothing about $P(A)$ itself, so (A) and (B) need not hold. (D) is impossible when $P(B) > 0$ ⁠, because $P(A \cap B) \le P(B)$ ⁠.

Why?

$A \cap B$ is part of $B$ ⁠, so its probability cannot be more than $P(B)$ ⁠. But $2P(B) > P(B)$ whenever $P(B) > 0$ ⁠.

Answer: (C) $P(A \cap B) = \dfrac{1}{2}\,P(B)$

Where this comes from

Chapter 13, Probability: Conditional probability.

Question type: Given numbers

How to spot it: click to see

Type 1. Given numbers: divide by the event after the bar

Spot it: the question gives $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, and asks for $P(E|F)$ or $P(F|E)$ ⁠.

Here: No numbers this time, but the same formula does it. $P(A|B) + P(A'|B) = 1$ makes each $\frac12$ ⁠; then write $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac12$ ⁠, with $B$ ⁠, the event after the bar, on the bottom.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q11: If P(A∣B)=P(A′∣B)P(A|B) = P(A'|B)P(A∣B)=P(A′∣B)⁠, then P(A∩B)=12P(B)P(A \cap B) = \frac{1}{2}P(B)P(A∩B)=21​P(B)

Q11: If $P(A|B) = P(A'|B)$ ⁠, then $P(A \cap B) = \frac{1}{2}P(B)$