JEE Advanced 2024 Paper 2 · Q09 · Probability

Question

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 ball drawn in the $i^{	ext{th}}$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P(W_{1} \cap G_{2} \cap B_{3}) = \dfrac{2}{5N}$ and the conditional probability $P(B_{3} \mid W_{1} \cap G_{2}) = \dfrac{2}{9}$, then $N$ equals ______.

SolveKar AI · Accepted Answer

Number of blue balls $= N - 9$. Then $P(W_{1} \cap G_{2} \cap B_{3}) = \dfrac{3}{N} \cdot \dfrac{6}{N-1} \cdot \dfrac{N-9}{N-2}$. Equating to $\dfrac{2}{5N}$ gives $\dfrac{18(N-9)}{(N-1)(N-2)} = \dfrac{2}{5}$, i.e. $45(N-9) = (N-1)(N-2)$. Also $P(B_{3} \mid W_{1} \cap G_{2}) = \dfrac{N-9}{N-2} = \dfrac{2}{9}$ gives $9(N-9) = 2(N-2) \Rightarrow 7N = 77 \Rightarrow N = 11$. (Verifies the first equation.) Hence $N = 11$.

JEE Advanced 2024 Paper 2 · Q09 · Probability

Solution