🎥 Video solution / Text Solution of this question is given below:
Given: Odds in favour of A = 2 : 1
So, $P(A) = \frac{2}{3}$
Odds in favour of $A \cup B = 3 : 1$
So, $P(A \cup B) = \frac{3}{4}$
Let $P(B) = q$ and $P(A \cap B) = x$.
Using inclusion–exclusion:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
So,
$\frac{3}{4} = \frac{2}{3} + q - x$
$\Rightarrow x = \frac{2}{3} + q - \frac{3}{4}$
Condition 1: $x \ge 0$
$\frac{2}{3} + q - \frac{3}{4} \ge 0$
$\Rightarrow q \ge \frac{1}{12}$
Condition 2: $x \le P(A)$
$\frac{2}{3} + q - \frac{3}{4} \le \frac{2}{3}$
$\Rightarrow q \le \frac{3}{4}$
Therefore possible range of $P(B)$ is:
$\frac{1}{12} \le P(B) \le \frac{3}{4}$
So minimum value = $\frac{1}{12}$
Maximum value = $\frac{3}{4}$