Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{1}{3}$

Given, $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{3}$ and $P\left(C\right)=\frac{1}{4}$
$\therefore P\left(A^{\prime }\right)=\frac{2}{3},P\left(B^{\prime }\right)=\frac{2}{3}$ and $P\left(C^{\prime }\right)=\frac{3}{4}$
Now, $P(A^{\prime }\cap B^{\prime }\cap C^{\prime })=P\left(A^{\prime }\right)P\left(B^{\prime }\right)P\left(C^{\prime }\right)$
($\because A,B$ and $C$ are mutually exclusive events)
$=\frac{2}{3}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{3}$