Find Total Balls N Using Conditional and Joint Probability

To find the total number of balls $N$, we analyze the composition of the bag and the given probabilities.

1. Composition of the Bag:

• Number of white balls = $3$
• Number of green balls = $6$
• Total number of balls = $N$
• Number of blue balls = $N - (3 + 6) = N - 9$

Since three balls are drawn sequentially without replacement:

• The probability of drawing a white ball in the first draw is: $P(W_1) = \frac{3}{N}$

• The conditional probability of drawing a green ball in the second draw, given that the first ball drawn was white, is: $P(G_2 \mid W_1) = \frac{6}{N-1}$

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

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2. Finding $N$ using the Conditional Probability: We are given that: $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$

Substituting the expression for $P(B_3 \mid W_1 \cap G_2)$: $\frac{N-9}{N-2} = \frac{2}{9}$

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

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3. Verification using the Joint Probability: By the multiplication rule of probability, the joint probability is: $P(W_1 \cap G_2 \cap B_3) = P(W_1 \cap G_2) \cdot P(B_3 \mid W_1 \cap G_2)$

Since $P(W_1 \cap G_2) = P(W_1) \cdot P(G_2 \mid W_1) = \frac{3}{N} \cdot \frac{6}{N-1} = \frac{18}{N(N-1)}$, we have: $P(W_1 \cap G_2 \cap B_3) = \frac{18}{N(N-1)} \cdot \frac{2}{9} = \frac{4}{N(N-1)}$

We are given that $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$. Equating the two expressions: $\frac{4}{N(N-1)} = \frac{2}{5N}$

Since $N > 0$, we can cancel $N$ from both denominators: $\frac{4}{N-1} = \frac{2}{5}$ $20 = 2(N - 1)$ $10 = N - 1 \implies N = 11$

Both conditions yield $N = 11$.

Thus, the value of $N$ is 11.