A bag contains 10 white and 15 red balls. If | vedclass

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(C) Total number of balls = $10 + 15 = 25$.Probability of drawing the first ball as red: $P(R_1) = \frac{15}{25} = \frac{3}{5}$.After drawing one red

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$1/4$

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(C) Total number of balls = $10 + 15 = 25$.Probability of drawing the first ball as red: $P(R_1) = \frac{15}{25} = \frac{3}{5}$.After drawing one red ball,the remaining number of balls is $24$,and the number of white balls remains $10$.Probability of drawing the second ball as white given the first was red: $P(W_2|R_1) = \frac{10}{24} = \frac{5}{12}$.The probability that the first is red and the second is white is: $P(R_1 \cap W_2) = P(R_1) 	imes P(W_2|R_1) = \frac{3}{5} 	imes \frac{5}{12} = \frac{15}{60} = \frac{1}{4}$.

$A$ bag contains $10$ white and $15$ red balls. If two balls are drawn one after another,what is the probability that the first ball is red and the second ball is white?

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