Question

$A$ and $B$ are two independent events such that $P\left(A\right)=\frac{1}{2};P\left(B\right)=\frac{1}{3}$. Then $P$(neither $A$ nor $B$) is equal to

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{5}{6}$
• D. $\frac{1}{3}$

Answer 1

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

Given $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3}$

Since $A,B$ are independent events $P(A\cap B)=P\left(A\right)\cdot P\left(B\right)$

$\therefore P(A\cap B)=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$\therefore P(A\cup B)=\frac{2}{3}$

$P\left(neitherAnorB\right)=P(A^{\prime }\cap B^{\prime })$

$=P(A\cup B{)}^{\prime }$

$=1-P(A\cup B)$

$=1-\frac{2}{3}$

$P\left(neitherAnorB\right)=\frac{1}{3}$