A box contains 3 white and 2 red balls. A bal | vedclass

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(B) The total number of balls is $3 + 2 = 5$.Let $W_1$ be the event that the first ball is white and $R_1$ be the event that the first ball is red.Let

Vedclass · Accepted Answer

$\frac{2}{5}$

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(B) The total number of balls is $3 + 2 = 5$.Let $W_1$ be the event that the first ball is white and $R_1$ be the event that the first ball is red.Let $R_2$ be the event that the second ball is red.The second ball can be red in two mutually exclusive ways:$(i)$ First ball is white and second ball is red: $P(W_1 \cap R_2) = P(W_1) 	imes P(R_2|W_1) = \frac{3}{5} 	imes \frac{2}{4} = \frac{6}{20}$.$(ii)$ First ball is red and second ball is red: $P(R_1 \cap R_2) = P(R_1) 	imes P(R_2|R_1) = \frac{2}{5} 	imes \frac{1}{4} = \frac{2}{20}$.The total probability of the second ball being red is $P(R_2) = P(W_1 \cap R_2) + P(R_1 \cap R_2) = \frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5}$.

$A$ box contains $3$ white and $2$ red balls. $A$ ball is drawn and another ball is drawn without replacing the first ball. Then,the probability of the second ball being red is:

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