Question

A and B are two events such that \(\rm P(A) = \dfrac{1}{2}\), \(\rm P(B) = \dfrac{1}{3}\) and \(\rm P(A\cap B) = \dfrac{1}{4}\). The values P(A|B) and P(B|A) are:

1. \(\rm \dfrac{1}{6},\ \dfrac{1}{6}\)
2. \(\rm \dfrac{2}{3},\ \dfrac{3}{2}\)
3. \(\rm \dfrac{3}{4},\ \dfrac{1}{2}\)
4. \(\rm \dfrac{3}{4},\ \dfrac{1}{4}\)
5. None of these

Answer 1

Correct Answer - Option 3 : \(\rm \dfrac{3}{4},\ \dfrac{1}{2}\)

Concept:

If A and B are two events, then the conditional probability of A given B is defined as:
\(\rm P(A|B)=\dfrac{P(A\cap B)}{P(B)}\), when P(B) > 0.

Solution:

Given that , \(\rm P(A) = \dfrac{1}{2}\), \(\rm P(B) = \dfrac{1}{3}\) and \(\rm P(A\cap B) = \dfrac{1}{4}\).

\(\rm P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\tfrac{1}{4}}{\tfrac{1}{3}}=\dfrac{1}{4}\times\dfrac{3}{1}=\dfrac{3}{4}\).

\(\rm P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{\tfrac{1}{4}}{\tfrac{1}{2}}=\dfrac{1}{4}\times\dfrac{2}{1}=\dfrac{1}{2}\).