A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue…
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 $P\left(W_1 \cap G_2 \cap B_3\right)=\frac{2}{5 N}$ and the conditional probability $P\left(B_3 \mid W_1 \cap G_2\right)=\frac{2}{9}$, then $N$ equals ________.
Given $\mathrm{P}\left(\mathrm{W}_1 \cap \mathrm{G}_2 \cap \mathrm{B}_3\right)=\frac{2}{5 \mathrm{~N}}$
and $\mathrm{P}\left(\mathrm{B}_3 \mid \mathrm{W}_1 \cap \mathrm{G}_2\right)=\frac{2}{9}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{P}\left(\mathrm{B}_3 \cap \mathrm{W}_1 \cap \mathrm{G}_2\right)}{\mathrm{P}\left(\mathrm{W}_1 \cap \mathrm{G}_2\right)}=\frac{2}{9} \\ & \Rightarrow \frac{2}{5 \mathrm{~N}} \times \frac{\mathrm{N} \times(\mathrm{N}-1)}{3 \times 6}=\frac{2}{9} \\ & \Rightarrow \mathrm{N}=11\end{aligned}$