Question

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

• A. $\frac{3}{4}$
• B. $\frac{5}{6}$
• C. $\frac{1}{4}$
• D. $\frac{3}{7}$

Answer 1

The correct option is C $\frac{1}{4}$

$P\left(\frac{B}{A\cup B^{\prime }}\right)=\frac{P(B\cap (A\cup B^{\prime }))}{P(A\cup B^{\prime })}=\frac{P((B\cap A)\cup (B\cap B^{\prime }))}{P\left(A\right)+P\left(B^{\prime }\right)-P(A\cap B^{\prime })}$
But $P((B\cap A)\cup (B\cap B^{\prime }))=P(B\cap A)=P\left(A\right)-P(A\cap B^{\prime })=0.2$
and $P\left(A\right)+P\left(B^{\prime }\right)-P(A\cap B^{\prime })=0.7+0.6-0.5=0.8$
$\therefore P\left(\frac{B}{A\cup B^{\prime }}\right)=\frac{0.2}{0.8}=\frac{1}{4}$