Question

If $X$ and $Y$ are two events that $P(\raisebox{1ex}{$X$}\!\left/ \!\raisebox{-1ex}{$Y$}\right.)=1/2,P\left(\raisebox{1ex}{$Y$}\!\left/ \!\raisebox{-1ex}{$X$}\right.\right)=1/3\mathrm{and}P(X\cap Y)==1/6$, then which of the following is/are correct?

• A. $P\left(X\cup Y\right)=\frac{2}{3}$
• B. $X$ and $Y$ are independent
• C. $X$ and $Y$ are not independent
• D. $P\left(X^c\cap Y\right)=\frac{1}{3}$

Answer 1

Answer: (A), (B)

$X$ and $Y$ are independent

Step 1: Explanation for the correct options.

Option A: $P\left(X\cup Y\right)=\frac{2}{3}$

$\begin{array}{rcl}P\left(\raisebox{1ex}{$X$}\!\left/ \!\raisebox{-1ex}{$Y$}\right.\right)& =& \frac{P\left(X\cap Y\right)}{P\left(Y\right)}\\ \Rightarrow \frac{1}{2}& =& \frac{{\displaystyle \frac{1}{6}}}{P\left(Y\right)}\\ \Rightarrow P\left(Y\right)& =& \frac{1}{3}\end{array}$

$\begin{array}{rcl}P\left(\raisebox{1ex}{$Y$}\!\left/ \!\raisebox{-1ex}{$X$}\right.\right)& =& \frac{P\left(X\cap Y\right)}{P\left(X\right)}\\ \Rightarrow \frac{1}{3}& =& \frac{{\displaystyle \frac{1}{6}}}{P\left(X\right)}\\ \Rightarrow P\left(X\right)& =& \frac{1}{2}\end{array}$

$\begin{array}{rcl}P\left(X\cup Y\right)& =& P\left(X\right)+P\left(Y\right)-P\left(X\cap Y\right)\\ & =& \frac{1}{2}+\frac{1}{3}-\frac{1}{6}\\ & =& \frac{3+2-1}{6}\\ & =& \frac{2}{3}\end{array}$

Therefore, option A is correct.

Option B: $X$ and $Y$ are independent.

If $X$ and $Y$ are independent, then

$\begin{array}{rcl}P\left(X\cap Y\right)& =& P\left(X\right)·P\left(Y\right)\\ \Rightarrow \frac{1}{6}& =& \frac{1}{2}\times \frac{1}{3}\\ \Rightarrow \frac{1}{6}& =& \frac{1}{6}\end{array}$

Therefore, option B is correct.

Step 2: Explanation for the incorrect options.

Option C: $X$ and $Y$ are not independent.

$\begin{array}{rcl}P\left(X\cap Y\right)& =& P\left(X\right)·P\left(Y\right)\end{array}$

Therefore, option C is incorrect.

Option D: $P\left(X^c\cap Y\right)=\frac{1}{3}$

$\begin{array}{rcl}P\left(X^c\cap Y\right)& =& P\left(Y\right)-P\left(X\cap Y\right)\\ \Rightarrow \frac{1}{3}& =& \frac{1}{3}-\frac{1}{6}\\ \Rightarrow \frac{1}{3}& \ne & \frac{1}{6}\end{array}$

Therefore, option D is incorrect.

Hence, option A and option B are correct.