Question

Let $X,Y$ be two events such that $P\left(X\right)=\frac{1}{2},P\left(X|Y\right)=\frac{2}{3},P\left(Y|X\right)=\frac{5}{6}$ then

• A. $P\left(Y\right)=\frac{5}{6}$
• B. $P\left(Y^{\prime }|X\right)=\frac{1}{6}$
• C. $P(X\cup Y)=\frac{17}{24}$
• D. $P(X\cap Y)=\frac{5}{12}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $P\left(Y^{\prime }|X\right)=\frac{1}{6}$
C $P(X\cup Y)=\frac{17}{24}$
D $P(X\cap Y)=\frac{5}{12}$

$P\left(X|Y\right)=\frac{P(X\cap Y)}{P\left(Y\right)}=\frac{2}{3}P\left(Y|X\right)=\frac{P(X\cap Y)}{P\left(X\right)}=\frac{5}{6}P(X\cap Y)=\frac{5}{12},P\left(Y\right)=\frac{\frac{5}{12}}{\frac{2}{3}}=\frac{5}{8}P\left(Y^{\prime }|X\right)=\frac{P(Y^{\prime }\cap X)}{P\left(X\right)}=\frac{P\left(X\right)-P(X\cap Y)}{P\left(X\right)}=\frac{1}{6}P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)=\frac{1}{2}+\frac{5}{8}-\frac{5}{12}=\frac{17}{24}$