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(C) Let $E_1$ be the event of selecting bag $X$ and $E_2$ be the event of selecting bag $Y$. Since the bag is selected at random, $P(E_1) = P(E_2) = \

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$\frac{8}{15}$

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(C) Let $E_1$ be the event of selecting bag $X$ and $E_2$ be the event of selecting bag $Y$. Since the bag is selected at random, $P(E_1) = P(E_2) = \frac{1}{2}$.Let $W$ be the event of drawing a white ball.The probability of drawing a white ball from bag $X$ is $P(W|E_1) = \frac{2}{2+3} = \frac{2}{5}$.The probability of drawing a white ball from bag $Y$ is $P(W|E_2) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.Using the law of total probability, the probability of drawing a white ball is $P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.$P(W) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

$A$ bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then, the probability for the ball chosen to be white is:

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