$\text{A bag contains}\mathrm{ N}\text{ balls out of which }3\text{ balls are white,}\mathrm{ 6}\text{ balls are green, and the remaining balls are blue.}$

$\text{Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement.}$

$\text{For}i=1,2,3,\text{ let}{\mathrm{ W}}_i,G_i,B_i\text{ denote the events that the ball drawn in the}{\mathrm{ i}}^{\text{th}}\text{draw is a white ball, green ball, and}$

$\text{blue ball, respectively. }\text{If the probability }P(W_1\cap G_2\cap B_3)=\frac{2}{5N}\text{ and the conditional probability}$

$P\left(B_3\right|W_1\cap G_2)=\frac{2}{9},$$\text{then}\mathrm{ N}\text{ equals \_\_\_\_\_}$

The correct answer is 11.

Step 1: Understand the composition of the bag

Let the total number of balls in the bag be N.

We are given the initial count of white and green balls:

• Number of white balls = 3
• Number of green balls = 6
• Number of blue balls = N - 3 - 6 = N - 9

Step 2: Formulate the conditional probability expression

The event $W_1\cap G_2$ indicates that a white ball is drawn on the first draw and a green ball is drawn on the second draw.

Since sampling is done without replacement, after drawing 1 white ball and 1 green ball, a total of 2 balls have been removed from the bag.

The remaining total number of balls in the bag before the third draw is:

$N-2$

Since no blue balls have been drawn yet, the number of remaining blue balls is still $N-9$.

Therefore, the conditional probability of drawing a blue ball on the 3^rd draw given that the 1^st was white and the 2^nd was green is given by:

$P\left(B_3\right|W_1\cap G_2)=\frac{N-9}{N-2}$

Step 3: Solve for N

We are given that this conditional probability equals $\frac{2}{9}$:

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

Cross-multiplying to solve for N:

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

Expanding both sides:

$9N-81=2N-4$

Rearranging terms to collect N on one side:

$9N-2N=81-4$

$7N=77$

$N=11$

Step 4: Verification using joint probability

We can double check using the given joint probability $P(W_1\cap G_2\cap B_3)=\frac{2}{5N}$:

$P(W_1\cap G_2\cap B_3)=P\left(W_1\right)·P\left(G_2\right|W_1)·P(B_3|W_1\cap G_2)$

For N = 11:

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

Evaluating $\frac{2}{5N}$ at N = 11 gives $\frac{2}{5\times 11}=\frac{2}{55}$, which matches perfectly.

Hence, the value of N equals 11.