If two events $A$ and $B$ are such that $P\left(A^{C}\right)=0.3$, $P(B)=0.4$ and $P\left(A \cap B^{

Question

If two events $A$ and $B$ are such that $P\left(A^{C}\right)=0.3$, $P(B)=0.4$ and $P\left(A \cap B^{C}\right)=0.5$ then $P\left(B /\left(A \cup B^{C}\right)\right)$ in % terms is

Professor PC Thomas Classes · Answer

$P\left(B /\left(A \cup B^{C}ight)ight)=\frac{P\left(B \cap\left(A \cup B^{C}ight)ight)}{P\left(A \cup B^{C}ight)}$\ $=\frac{P\left((B \cap A) \cup\left(B \cap B^{C}ight)ight)}{P\left(A \cup B^{C}ight)}$\ $=\frac{P(A \cap B)}{P\left(A \cup B^{C}ight)}=\frac{P(A)-P\left(A \cap B^{C}ight)}{P\left(A \cup B^{C}ight)}$\ $=\frac{(1-0.3)-0.5}{(1-0.3)+(1-0.4)-0.5}=0.25$.\