An urn contains 4 red and 5 white balls. Two | vedclass

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(B) Total number of red balls $= 4$. Total number of white balls $= 5$. Total number of balls $= 4 + 5 = 9$. Let $R_1$ be the event that the first bal

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$\frac{1}{6}$

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(B) Total number of red balls $= 4$. Total number of white balls $= 5$. Total number of balls $= 4 + 5 = 9$. Let $R_1$ be the event that the first ball drawn is red and $R_2$ be the event that the second ball drawn is red. The probability of drawing the first red ball is $P(R_1) = \frac{4}{9}$. Since the balls are drawn without replacement,if the first ball is red,there are now $3$ red balls left out of $8$ total balls. The probability of drawing the second red ball given that the first was red is $P(R_2|R_1) = \frac{3}{8}$. The probability that both balls are red is $P(R_1 \cap R_2) = P(R_1) 	imes P(R_2|R_1) = \frac{4}{9} 	imes \frac{3}{8} = \frac{12}{72} = \frac{1}{6}$.

An urn contains $4$ red and $5$ white balls. Two balls are drawn one after the other without replacement,then the probability that both the balls are red is

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