Probability
PYP_JEE_ADV_2024_P2
Grade None
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 $i^{\text{th}}$ draw is a white ball, green ball, and blue ball, respectively. If the probability
$$P(W_1 \cap G_2 \cap B_3) = \dfrac{2}{5N} \quad \text{and the conditional probability} \quad P(B_3 \mid W_1 \cap G_2) = \dfrac{2}{9},$$
then $N$ equals ___.
Step-by-Step Solution
Key Concept: Multiplication theorem of probability for dependent events and conditional probability of drawing balls without replacement.
Let the number of blue balls be $b = N - 9$.
Drawing without replacement, after drawing one white and one green ball, the total remaining balls in the bag is $N-2$, and the number of remaining blue balls is $b = N-9$.
The conditional probability of drawing a blue ball on the third draw is:
$$P(B_3 \mid W_1 \cap G_2) = \dfrac{N-9}{N-2}$$
We are given this is $\dfrac{2}{9}$:
$$\dfrac{N-9}{N-2} = \dfrac{2}{9} \implies 9(N-9) = 2(N-2) \implies 7N = 77 \implies N = 11$$
Let's verify with the second given condition:
$$P(W_1 \cap G_2 \cap B_3) = P(W_1) \cdot P(G_2 \mid W_1) \cdot P(B_3 \mid W_1 \cap G_2)$$
$$= \dfrac{3}{11} \cdot \dfrac{6}{10} \cdot \dfrac{2}{9} = \dfrac{36}{990} = \dfrac{2}{55}$$
Since $N = 11$, $\dfrac{2}{5N} = \dfrac{2}{55}$, which matches perfectly.
Thus, the answer is 11.
Correct Answer: 11