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^ th draw is a white ball, green ball, and blue ball, respectively. If the probability P(W_1 \cap G_2 \cap B_3)=\frac25 N and the conditional probability P(B_3 \mid W_1 \cap G_2)=\frac29 , then N equals \underline\hspace2cm .

Probability PYQ — JEE Advanced 2024 · Shift 2 · Q26 with Solution

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Correct answer: 11

Let the total number of balls be $N$ .

Since three balls are drawn without replacement, we use sequential probabilities.

We are given

P(B_3\mid W_1\cap G_2)=\frac{2}{9}.

If the first draw is white and the second is green, then two balls are already removed:

Hence,

P(B_3\mid W_1\cap G_2)=\frac{N-9}{N-2}.

So,

\frac{N-9}{N-2}=\frac{2}{9}.

Cross-multiplying,

9(N-9)=2(N-2)

9N-81=2N-4

7N=77

N=11.

Now check whether this satisfies

P(W_1\cap G_2\cap B_3)=\frac{2}{5N}.

In general,

P(W_1\cap G_2\cap B_3)=P(W_1)P(G_2\mid W_1)P(B_3\mid W_1\cap G_2).

Compute each factor:

P(W_1)=\frac{3}{N},

P(G_2\mid W_1)=\frac{6}{N-1},

P(B_3\mid W_1\cap G_2)=\frac{N-9}{N-2}.

Thus,

P(W_1\cap G_2\cap B_3)=\frac{3}{N}\cdot \frac{6}{N-1}\cdot \frac{N-9}{N-2}.

For $N=11$ ,

P(W_1\cap G_2\cap B_3)=\frac{3}{11}\cdot \frac{6}{10}\cdot \frac{2}{9} =\frac{36}{990}=\frac{2}{55}.

And

\frac{2}{5N}=\frac{2}{5\cdot 11}=\frac{2}{55}.

This matches perfectly.

\boxed{11}