1 bag contains 4 white and 2 black balls. Ano | vedclass

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(C) Let $E_1$ be the event of drawing a white ball from the first bag.Total balls in the first bag $= 4 + 2 = 6$.Probability $P(E_1) = \frac{4}{6} = \

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) Let $E_1$ be the event of drawing a white ball from the first bag.Total balls in the first bag $= 4 + 2 = 6$.Probability $P(E_1) = \frac{4}{6} = \frac{2}{3}$.Let $E_2$ be the event of drawing a white ball from the second bag.Total balls in the second bag $= 3 + 5 = 8$.Probability $P(E_2) = \frac{3}{8}$.Since the events are independent,the probability that both balls are white is $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$.$P(E_1 \cap E_2) = \frac{2}{3} 	imes \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

$1$ bag contains $4$ white and $2$ black balls. Another bag contains $3$ white and $5$ black balls. If $1$ ball is drawn from each bag,find the probability that both balls are white.

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