If $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $

Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$

Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$

Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$

The probabilities of three mutually exclusive events are $\frac{2}{3} ,  \frac{1}{4}$ and $\frac{1}{6}$. The statement is

If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that $2$ white and $1$ black ball will be drawn is

If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $