Question

If two events $A$ and $B$ are such that $P\left(A^{\prime }\right)=0.3,P\left(B\right)=0.4$ and $P\left(A{B}^{\prime }\right)=0.5$, then $P\left(\frac{B}{(A\cup B^{\prime })}\right)=$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{4}$

$P\left(A\right)=1-P\left(A^{\prime }\right)=1-0.3=0.7,P\left(B\right)=0.4$
We know that
$P\left(A\right)=P(A\cap B)+P(A\cap B^{\prime })\Rightarrow 0.7=P(A\cap B)=0.5\Rightarrow P(A\cap B)=0.7-0.5=0.2P(A\cup B^{\prime })=P\left(A\right)+P\left(B^{\prime }\right)-P(A\cap B^{\prime })=0.7+(1-0.4)-0.5=0.8$
Next $P\left(\frac{B}{(A\cup B^{\prime })}\right)=\frac{P[B\cap (A\cap B^{\prime })]}{P(A\cap B^{\prime })}$
From set theory $B\cap (A\cap B^{\prime })=(B\cap A)\cup (B\cap B^{\prime })=(B\cap A)\cup \varphi =B\cap A$
$\therefore$ Required probability $=\frac{P(A\cap B)}{P(A\cup B^{\prime })}=\frac{0.2}{0.8}=\frac{1}{4}$