A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i = 1, 2, 3$, let $W_i, G_i$, and $B_i$ denote the events that the...

The correct numerical answer is 11. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Mathematics Question 9 Solution

Let $b$ be the number of blue balls. Then total balls $N = 3 + 6 + b = 9 + b$ .
The probability $P(W_1 \cap G_2 \cap B_3) = \frac{3}{N} \cdot \frac{6}{N-1} \cdot \frac{b}{N-2} = \frac{18b}{N(N-1)(N-2)}$ .
Given $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ , we have $\frac{18b}{N(N-1)(N-2)} = \frac{2}{5N} \implies \frac{9b}{(N-1)(N-2)} = \frac{1}{5} \implies 45b = (N-1)(N-2)$ .
The conditional probability $P(B_3 | W_1 \cap G_2) = \frac{P(W_1 \cap G_2 \cap B_3)}{P(W_1 \cap G_2)}$ . Since after drawing $W_1$ and $G_2$ , there are $N-2$ balls remaining and $b$ are blue, $P(B_3 | W_1 \cap G_2) = \frac{b}{N-2}$ .
Given $\frac{b}{N-2} = \frac{2}{9} \implies b = \frac{2(N-2)}{9}$ .
Substitute $b$ into the first equation: $45 \left(\frac{2(N-2)}{9}\right) = (N-1)(N-2) \implies 10(N-2) = (N-1)(N-2)$ .
Since $b > 0$ , $N > 2$ , so $10 = N - 1 \implies N = 11$ .