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 and W_i G_i 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 $P(W_1\cap <!--\; \cap \; -->G_2\cap <!--\; \cap \; -->B_3)=\frac{2}{5N}$ and the conditional probability $P(B_3\mid <!--\; \mid \; -->W_1\cap <!--\; \cap \; -->G_2)=\frac{2}{9}$, then N equals ______.

The correct answer is 11.

Let's solve the problem step-by-step.

We are given a bag containing a total of $N$ balls, which consists of:
- White balls: 3
- Green balls: 6
- Blue balls: $N-9$

Three balls are drawn randomly one after another without replacement. Let $W_1$ be the event that the first ball drawn is white, $G_2$ be the event that the second ball drawn is green, and $B_3$ be the event that the third ball drawn is blue.

We are given the conditional probability:
$P(B_3\mid W_1\cap G_2)=\frac{2}{9}$

Let us express the conditional probability in terms of $N$. After drawing one white ball in the first draw and one green ball in the second draw:
- The number of remaining balls in the bag is $N-2$.
- The number of remaining blue balls is still $N-9$ (since no blue ball has been drawn yet).

Therefore, the conditional probability of drawing a blue ball on the third draw, given that the first was white and the second was green, is:
$P(B_3\mid W_1\cap G_2)=\frac{N-9}{N-2}$

Equating this to the given value:
$\frac{N-9}{N-2}=\frac{2}{9}$

Cross-multiplying to solve for $N$:
$9(N-9)=2(N-2)$
$9N-81=2N-4$
$7N=77$
$N=11$

We can verify this value using the second probability condition given in the problem, namely:
$P(W_1\cap G_2\cap B_3)=\frac{2}{5N}$

By the multiplication rule of probability, we have:
$P(W_1\cap G_2\cap B_3)=P\left(W_1\right)\cdot P(G_2\mid W_1)\cdot P(B_3\mid W_1\cap G_2)$

Substituting the corresponding probabilities:
- $P\left(W_1\right)=\frac{3}{N}$
- $P(G_2\mid W_1)=\frac{6}{N-1}$
- $P(B_3\mid W_1\cap G_2)=\frac{2}{9}$

Thus:
$P(W_1\cap G_2\cap B_3)=\frac{3}{N}\cdot \frac{6}{N-1}\cdot \frac{2}{9}=\frac{36}{9N(N-1)}=\frac{4}{N(N-1)}$

Equating this to the given probability value:
$\frac{4}{N(N-1)}=\frac{2}{5N}$

Since $N\ne 0$, we can divide both sides by $N$:
$\frac{4}{N-1}=\frac{2}{5}$
$2(N-1)=20$
$N-1=10$
$N=11$

Both equations yield the same consistent value. Thus, $N=11$.