A bag B contains 4 white balls and 2 black ba | vedclass

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(A) bag $B$ contains $4$ white balls and $2$ black balls. Total balls in bag $B = 4 + 2 = 6$. Probability of selecting $1$ white ball from bag $B$ is

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) bag $B$ contains $4$ white balls and $2$ black balls. Total balls in bag $B = 4 + 2 = 6$. Probability of selecting $1$ white ball from bag $B$ is $P_1 = \frac{4}{6} = \frac{2}{3}$. Another bag $C$ contains $3$ white balls and $5$ black balls. Total balls in bag $C = 3 + 5 = 8$. Probability of selecting $1$ white ball from bag $C$ is $P_2 = \frac{3}{8}$. Since the events are independent, the probability that both balls drawn are white is $P = P_1 	imes P_2 = \frac{2}{3} 	imes \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

$A$ bag $B$ contains $4$ white balls and $2$ black balls. Another bag $C$ contains $3$ white balls and $5$ black balls. If one ball is drawn randomly from each bag, then the probability that the two balls drawn are both white is

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