A bag contains N balls out of which 3 are whi | vedclass

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(D) Given that there are $3$ white,$6$ green,and $(N-9)$ blue balls in a bag of $N$ balls.The probability of drawing a white ball first,a green ball s

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$11$

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(D) Given that there are $3$ white,$6$ green,and $(N-9)$ blue balls in a bag of $N$ balls.The probability of drawing a white ball first,a green ball second,and a blue ball third without replacement is:$P(W_1 \cap G_2 \cap B_3) = P(W_1) 	imes P(G_2 \mid W_1) 	imes P(B_3 \mid W_1 \cap G_2)$We are given $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and $P(B_3 \mid W_1 \cap G_2) = \frac{2}{9}$.Substituting the values:$P(W_1) = \frac{3}{N}$$P(G_2 \mid W_1) = \frac{6}{N-1}$$P(B_3 \mid W_1 \cap G_2) = \frac{N-9}{N-2}$Thus,$\frac{3}{N} 	imes \frac{6}{N-1} 	imes \frac{N-9}{N-2} = \frac{2}{5N}$.Also,from the definition of conditional probability:$P(B_3 \mid W_1 \cap G_2) = \frac{N-9}{N-2} = \frac{2}{9}$.Solving for $N$:$9(N-9) = 2(N-2)$$9N - 81 = 2N - 4$$7N = 77$$N = 11$.

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