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$
$\therefore P[B/(A \cup B^C)]=\frac{P\{B \cap (A \cup B^c)\})}{P(A \cup B^c)}$
$=\frac{P\{(B \cap A)\cup (B \cap B^C)\}}{0.8}=\frac{P\{(B \cap A)\cup \Phi\}}{0.8}=\frac{P(B \cap A)}{0.8}$
=$\frac{1}{0.8}[P(A)-P(A \cap B^c)]=\frac{0.7-0.5}{0.8}=\frac{0.2}{0.8}=\frac{1}{4}$