Question

If two events $A$ and $B$ are such that $P(A^C)=0.3,P(B)=0.4$ and $P(A\cap B^C)=0.5$, then $P[B/(A\cup B^C)]=$....

Answer 1

$P(A^c)=0.3$ [given]
$\Rightarrow P(A)=0.7$
$P(B)=0.4$ [given]
$\Rightarrow P(B^c)=0.6$ and $P(A\cap B^c)=0.5$ [given]
Now, $P(A\cup B^c)=P(A)+P(B^c)-P(A\cap B^c)$
$=0.7+0.6-0.5=0.8$
$\therefore P[B/(A\cup B^C)]=\frac{P\{B\cap (A\cup B^c)\})}{P(A\cup B^c)}$
$=\frac{P\{(B\cap A)\cup (B\cap B^C)\}}{0.8}=\frac{P\{(B\cap A)\cup \Phi \}}{0.8}=\frac{P(B\cap A)}{0.8}$
=$\frac{1}{0.8}[P(A)-P(A\cap B^c)]=\frac{0.7-0.5}{0.8}=\frac{0.2}{0.8}=\frac{1}{4}$