Explanation

(a)
Given $A$ and $B$ are two events odds against $A$ are 2 to 1, odds in favour of $A \cup B$ are 3 to 1 also $x \leq P(B) \leq y$
Given $P(A) = \frac{1}{3},\ P(A \cup B) = \frac{3}{4}$
$\therefore\ P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\Rightarrow \ \frac{1}{3} + P(B) - P(A \cap B) = \frac{3}{4}$
$\Rightarrow \ P(A \cap B) = P(B) - \frac{5}{12}$
$\Rightarrow \ P(B) \geq \frac{5}{12}$
Again, $P(B) = \frac{5}{12} + P(A \cap B)$
$\Rightarrow \ P(B) \leq \frac{5}{12} + P(A)\ \ \lbrack\because P(A \cap B) \leq P(A)\rbrack$
$\Rightarrow \ P(B) \leq \frac{5}{12} + \frac{1}{3}\ \Rightarrow \ P(B) \leq \frac{3}{4}$
$\therefore\ \frac{5}{12} \leq P(B) \leq \frac{3}{4}$
But$x \leq P(B) \leq y$