Let $X$ and $Y$ be two events such that $P(X|Y)=\frac{1}{2}$, $P(Y|X)=\frac{1}{3}$ and $P(X\cap Y)=\frac{1}{6}$. Which of the following is(are) correct?

$\text{Given: } P(X|Y) = \dfrac{1}{2},\quad P(Y|X) = \dfrac{1}{3}$ $P(X \cap Y) = \dfrac{1}{6}$ $\text{To find: which of the following are correct?}$ $(1)\ P(X \cup Y) = \dfrac{2}{3} \qquad (2)\ X, Y \text{ independent}$ $(3)\ X, Y \text{ not independent} \qquad (4)\ P(X^c \cap Y) = \dfrac{1}{3}$ $P(X \cap Y) = \dfrac{1}{6} \ \text{(given)}$ $\text{Plan: } 1.\ P(Y) \text{ from } P(X|Y)$ $2.\ P(X) \text{ from } P(Y|X)$ $3.\ \text{test independence and each option}$ $P(X|Y) = \dfrac{P(X \cap Y)}{P(Y)}$ $P(Y|X) = \dfrac{P(X \cap Y)}{P(X)}$ $\dfrac{1}{2} = \dfrac{1/6}{P(Y)} \Rightarrow P(Y) = \dfrac{1}{6} \times 2 = \dfrac{1}{3}$ $\dfrac{1}{3} = \dfrac{1/6}{P(X)} \Rightarrow P(X) = \dfrac{1}{6} \times 3 = \dfrac{1}{2}$ $P(X) = \dfrac{1}{2},\quad P(Y) = \dfrac{1}{3},\quad P(X \cap Y) = \dfrac{1}{6}$ $P(X \text{ only}) = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{1}{3}$ $P(Y \text{ only}) = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$ $P(X)\,P(Y) = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6} = P(X \cap Y)$ $\Rightarrow X, Y \text{ are independent} \Rightarrow (2) \text{ true},\ (3) \text{ false}$ $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ $= \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{2}{3} \Rightarrow (1) \text{ true}$ $P(X^c \cap Y) = P(Y) - P(X \cap Y) = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$ $\dfrac{1}{6} \ne \dfrac{1}{3} \Rightarrow (4) \text{ is false}$ $\therefore (1) \text{ and } (2) \text{ are correct}$