Question

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 \( P(B_3 \mid W_1 \cup G_2) = \frac{2}{9} \), then \( N \) equals \_\_\_\_\_.}

Hint:

Break down conditional probabilities using known values to form solvable equations.

Answer 1

Given: \[ P(W_1 \cap G_2 \cap B_3) = \frac{3}{N} \cdot \frac{6}{N-1} \cdot \frac{N-9}{N-2} = \frac{2}{5N}. \] Simplifying: \[ \frac{18(N-9)}{N(N-1)(N-2)} = \frac{2}{5N}. \] Cross-multiplying: \[ 90(N-9) = 2N(N-1)(N-2). \] Expanding and simplifying: \[ N^2 - 48N + 407 = 0. \] Roots: \[ N = 11 \quad \text{or} \quad N = 37 \quad (\text{but \( N = 37 \) is invalid since \( N<15 \)}). \] Thus, \( N = 11 \).