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 $$P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$$ and the conditional probability $$P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$$, then $$N$$ equals ______.

Let the total number of balls be $$N$$.
Numbers of each colour:
    White   = $$3$$
    Green  = $$6$$
    Blue   = $$N-9$$

Because the draws are without replacement, multiply the successive probabilities.

Step 1 : Compute $$P(W_1 \cap G_2 \cap B_3)$$
$$P(W_1)=\frac{3}{N}$$
After removing one white ball, total balls become $$N-1$$ and green balls remain $$6$$, so
$$P(G_2 \mid W_1)=\frac{6}{N-1}$$
After removing a white and a green ball, total balls become $$N-2$$ and blue balls are still $$N-9$$, so
$$P(B_3 \mid W_1 \cap G_2)=\frac{N-9}{N-2}$$
Therefore
$$P(W_1 \cap G_2 \cap B_3)=\frac{3}{N}\cdot\frac{6}{N-1}\cdot\frac{N-9}{N-2}=\frac{18(N-9)}{N(N-1)(N-2)}$$

This probability is given to be $$\dfrac{2}{5N}$$, hence
$$\frac{18(N-9)}{N(N-1)(N-2)}=\frac{2}{5N} \quad -(1)$$

Step 2 : Compute the conditional probability $$P(B_3 \mid W_1 \cap G_2)$$
First find $$P(W_1 \cap G_2)$$:
$$P(W_1 \cap G_2)=\frac{3}{N}\cdot\frac{6}{N-1}=\frac{18}{N(N-1)}$$
Thus
$$P(B_3 \mid W_1 \cap G_2)=\frac{P(W_1 \cap G_2 \cap B_3)}{P(W_1 \cap G_2)} =\frac{\dfrac{18(N-9)}{N(N-1)(N-2)}}{\dfrac{18}{N(N-1)}} =\frac{N-9}{N-2}$$
This conditional probability is given to be $$\dfrac{2}{9}$$, so
$$\frac{N-9}{N-2}=\frac{2}{9} \quad -(2)$$

Step 3 : Solve equation $$(2)$$ for $$N$$
Cross-multiplying:
$$9(N-9)=2(N-2)$$
$$9N-81=2N-4$$
$$7N=77$$
$$N=11$$

Step 4 : Verification with equation $$(1)$$ (optional but reassuring)
Substitute $$N=11$$ in $$(1)$$:
Left side $$=\dfrac{18(11-9)}{11\cdot10\cdot9}=\dfrac{18\cdot2}{990}=\dfrac{36}{990}=\dfrac{2}{55}$$
Right side $$=\dfrac{2}{5\cdot11}=\dfrac{2}{55}$$
Both sides match, confirming the value.

Hence the total number of balls is
11