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(A) Let $E_1$ be the event of drawing a white ball from the first bag and $E_2$ be the event of drawing a white ball from the second bag.The first bag

Vedclass · Accepted Answer

$\frac{1}{4}$

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(A) Let $E_1$ be the event of drawing a white ball from the first bag and $E_2$ be the event of drawing a white ball from the second bag.The first bag contains $4$ white balls and $2$ black balls,so the total number of balls is $4 + 2 = 6$.The probability of drawing a white ball from the first bag is $P(E_1) = \frac{4}{6} = \frac{2}{3}$.The second bag contains $3$ white balls and $5$ black balls,so the total number of balls is $3 + 5 = 8$.The probability of drawing a white ball from the second bag is $P(E_2) = \frac{3}{8}$.Since the events are independent,the probability that both balls are white is $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$.$P(E_1 \cap E_2) = \frac{2}{3} 	imes \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

One bag contains $4$ white balls and $2$ black balls; another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag,find the probability that both are white.

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