There are two boxes, each containing 10 balls | vedclass

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(C) Let $B_1$ and $B_2$ be the events of selecting the first and second box respectively. Since the box is chosen at random, $P(B_1) = P(B_2) = \frac{

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$4$ or $8$

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(C) Let $B_1$ and $B_2$ be the events of selecting the first and second box respectively. Since the box is chosen at random, $P(B_1) = P(B_2) = \frac{1}{2}$.Let $E$ be the event of drawing a black ball.Let $n_1$ and $n_2$ be the number of black balls in the first and second box respectively. Since each box has $10$ balls, $P(E|B_1) = \frac{n_1}{10}$ and $P(E|B_2) = \frac{n_2}{10}$.By Bayes' Theorem, the probability that the ball is from the second box given it is black is:$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2)} = \frac{\frac{1}{2} 	imes \frac{n_2}{10}}{\frac{1}{2} 	imes \frac{n_1}{10} + \frac{1}{2} 	imes \frac{n_2}{10}} = \frac{n_2}{n_1 + n_2}$.Given $P(B_2|E) = \frac{1}{5}$, we have $\frac{n_2}{n_1 + n_2} = \frac{1}{5}$, which implies $5n_2 = n_1 + n_2$, or $n_1 = 4n_2$.Since $n_1$ and $n_2$ are integers between $0$ and $10$, we test possible values for $n_2$:If $n_2 = 1$, $n_1 = 4$.If $n_2 = 2$, $n_1 = 8$.If $n_2 = 0$, $n_1 = 0$ (not possible as a black ball was drawn).Thus, $n_1$ can be $4$ or $8$.

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