Question

$A$ and $B$ are two independent events such that $P(A\cup B^{\prime })=0.8$, and $P\left(A\right)=0.3$. Then, $P\left(B\right)$ is

• A. $\frac{2}{7}$
• B. $\frac{2}{3}$
• C. $\frac{3}{8}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{2}{7}$

$P(A\cup B^{\prime })=P\left(A\right)+P\left(B^{\prime }\right)-P\left(A\right)P\left(B^{\prime }\right)$
$\therefore 0.8=0.3+P\left(B^{\prime }\right)-0.3P\left(B^{\prime }\right)$
$\Rightarrow 0.5=P\left(B^{\prime }\right)\left(0.7\right)$
$\Rightarrow P\left(B^{\prime }\right)=\frac{5}{7}$
$\therefore P\left(B\right)=1-\frac{5}{7}=\frac{2}{7}$