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 $P(W_{1} ∩ G_{2} ∩ B_{3}) = 2/(5N)$ and the conditional probability $P(B_{3} | W_{1} ∩ G_{2}) = \frac{2}{9},$ then N equals ______.

$\text{White} = 3,\quad \text{Green} = 6,\quad \text{Blue} = N - 9,\quad \text{Total} = N$ $3 \text{ draws, randomly, without replacement}$ $P(W_1 \cap G_2 \cap B_3) = \dfrac{2}{5N}$ $P(B_3 \mid W_1 \cap G_2) = \dfrac{2}{9}$ $\text{To find: } N$ $P(W_1) = \dfrac{3}{N},\quad P(G_1) = \dfrac{6}{N},\quad P(B_1) = \dfrac{N-9}{N}$ $P(W_2 \mid W_1) = \dfrac{2}{N-1},\quad P(G_2 \mid W_1) = \dfrac{6}{N-1},\quad P(B_2 \mid W_1) = \dfrac{N-9}{N-1}$ $P(W_3 \mid W_1 \cap G_2) = \dfrac{2}{N-2},\quad P(G_3 \mid W_1 \cap G_2) = \dfrac{5}{N-2}$ $P(B_3 \mid W_1 \cap G_2) = \dfrac{N-9}{N-2}$ $1.\ P(W_1 \cap G_2 \cap B_3) \text{ by the multiplication rule}$ $2.\ \text{set it equal to } \dfrac{2}{5N} \Rightarrow \text{a quadratic in } N$ $3.\ \text{use } P(B_3 \mid W_1 \cap G_2) = \dfrac{2}{9} \text{ to select the right } N$ $P(W_1 \cap G_2 \cap B_3) = P(W_1)\,P(G_2 \mid W_1)\,P(B_3 \mid W_1 \cap G_2)$ $= \dfrac{3}{N} \cdot \dfrac{6}{N-1} \cdot \dfrac{N-9}{N-2} = \dfrac{18(N-9)}{N(N-1)(N-2)}$ $\dfrac{18(N-9)}{N(N-1)(N-2)} = \dfrac{2}{5N}$ $\Rightarrow \dfrac{18(N-9)}{(N-1)(N-2)} = \dfrac{2}{5}$ $\Rightarrow 18(N-9)\cdot 5 = 2(N-1)(N-2)$ $\Rightarrow 45(N-9) = (N-1)(N-2)$ $45N - 405 = N^2 - 3N + 2$ $\Rightarrow N^2 - 48N + 407 = 0$ $D = 48^2 - 4(407) = 2304 - 1628 = 676 = 26^2$ $N = \dfrac{48 \pm 26}{2} = 11 \ \text{or} \ 37$ $P(B_3 \mid W_1 \cap G_2) = \dfrac{N-9}{N-2} = \dfrac{2}{9}$ $9(N-9) = 2(N-2)$ $9N - 81 = 2N - 4 \Rightarrow 7N = 77 \Rightarrow N = 11$ $N = 37:\ \dfrac{37-9}{37-2} = \dfrac{28}{35} = \dfrac{4}{5} \neq \dfrac{2}{9}\ \text{✗}$ $\Rightarrow N = 11$ $N = 11:\ 3W,\ 6G,\ 2B$ $P(W_1 \cap G_2 \cap B_3) = \dfrac{3}{11}\cdot\dfrac{6}{10}\cdot\dfrac{2}{9} = \dfrac{36}{990} = \dfrac{2}{55}$ $\dfrac{2}{5N} = \dfrac{2}{55}\ \checkmark$ $P(B_3 \mid W_1 \cap G_2) = \dfrac{11-9}{11-2} = \dfrac{2}{9}\ \checkmark$ $\boxed{N = 11}$