A bag contains 3 black and 4 white balls. Two | vedclass

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(C) Let $W_1$ be the event that the first ball is white and $B_1$ be the event that the first ball is black. Let $W_2$ be the event that the second ba

Vedclass · Accepted Answer

$\frac{4}{7}$

Vedclass · Answer

(C) Let $W_1$ be the event that the first ball is white and $B_1$ be the event that the first ball is black. Let $W_2$ be the event that the second ball is white.The second ball is white in two mutually exclusive cases:$1$. The first ball is white and the second ball is white $(W_1 \cap W_2)$:$P(W_1 \cap W_2) = P(W_1) 	imes P(W_2|W_1) = \frac{4}{7} 	imes \frac{3}{6} = \frac{12}{42} = \frac{2}{7}$.$2$. The first ball is black and the second ball is white $(B_1 \cap W_2)$:$P(B_1 \cap W_2) = P(B_1) 	imes P(W_2|B_1) = \frac{3}{7} 	imes \frac{4}{6} = \frac{12}{42} = \frac{2}{7}$.The total probability that the second ball is white is $P(W_2) = P(W_1 \cap W_2) + P(B_1 \cap W_2) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$.

$A$ bag contains $3$ black and $4$ white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white,is

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