If two events $A$ and $B$ are such that $P(A')=0.3$, $P(B)=0.5$ and $P(A\cap B)=0.3$, then $P(B\mid(A\cup B'))$ is

Explanation

Given $P(A')=0.3$.\n\nSo, $P(A)=1-0.3=0.7$.\n\nAlso, $P(B)=0.5$, hence $P(B')=1-0.5=0.5$.\n\nNow,\n\n$P(A\cap B')=P(A)-P(A\cap B)=0.7-0.3=0.4$.\n\nTherefore,\n\n$P(A\cup B')=P(A)+P(B')-P(A\cap B')$\n\n$=0.7+0.5-0.4=0.8$.\n\nNow,\n\n$P(B\mid(A\cup B'))=\frac{P(B\cap(A\cup B'))}{P(A\cup B')}$\n\n$=\frac{P((B\cap A)\cup(B\cap B'))}{0.8}$\n\n$=\frac{P(A\cap B)}{0.8}$\n\n$=\frac{0.3}{0.8}$\n\n$=\frac{3}{8}$.\n\nHence the answer is $\frac{3}{8}$