Question

If A, B and C are three events such that $P\left(B\right)=\frac{3}{4},P(A\cap B\cap C^{\prime })=\frac{1}{3}$ and $P(A^{\prime }\cap B\cap C^{\prime })=\frac{1}{3}$, then $P(B\cap C)$ is equal to

• A. $\frac{1}{12}$
• B. $\frac{1}{6}$
• C. $\frac{1}{15}$
• D. $\frac{1}{9}$

Answer 1

The correct option is A $\frac{1}{12}$

We have
$P(B\cap C^{\prime })=P\left[\right(A\cap A^{\prime })\cap (B\cap C^{\prime }\left)\right]$
$=P(A\cap B\cap C^{\prime })+P(A^{\prime }\cap B\cap C^{\prime })=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$
Now, $P(B\cap C)=P\left(B\right)-P(B\cap C^{\prime })=\frac{3}{4}-\frac{2}{3}=\frac{1}{12}$