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

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Question

(B) 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

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(B) 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 cases:Case $(i)$: The first ball is white and the second ball is red.$P(W_1 \cap R_2) = P(W_1) 	imes P(R_2 | W_1) = \frac{3}{5} 	imes \frac{2}{4} = \frac{6}{20}$.Case $(ii)$: The first ball is red and the 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 that the second ball is 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. What is the probability that the second ball is red?

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