Q.It is given that events $A$ and $B$ are such that $P(A) = \dfrac{1}{4}$, $P(A/B) = \dfrac{1}{2}$ and $P(B/A) = \dfrac{2}{3}$. Then $P(B)$ is

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$. But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$, and two of those (4 and 6) are even, so the probability becomes $2/3$. That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

The Definition

For events $A$ and $B$ with $P(B) > 0$,

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$; the surviving part of $A$ is the overlap $A \cap B$. Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$, so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$, matching the intuition.

Two Cautions

• $P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$. …