A bag contains 3 red,6 white,and 7 blue balls | vedclass

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(D) Total number of balls = $3 + 6 + 7 = 16$.Let $A$ be the event that the first ball is white.Let $B$ be the event that the second ball is blue.Since

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$7/40$

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(D) Total number of balls = $3 + 6 + 7 = 16$.Let $A$ be the event that the first ball is white.Let $B$ be the event that the second ball is blue.Since the first ball is not replaced,the events are dependent.The probability of drawing a white ball first is $P(A) = \frac{6}{16} = \frac{3}{8}$.After drawing one white ball,the remaining number of balls in the bag is $15$.The number of blue balls remains $7$.The conditional probability of drawing a blue ball given that the first was white is $P(B|A) = \frac{7}{15}$.The probability that the first ball is white and the second is blue is $P(A \cap B) = P(A) 	imes P(B|A) = \frac{6}{16} 	imes \frac{7}{15} = \frac{3}{8} 	imes \frac{7}{15} = \frac{1}{8} 	imes \frac{7}{5} = \frac{7}{40}$.

$A$ bag contains $3$ red,$6$ white,and $7$ blue balls. Two balls are drawn one after another. What is the probability that the first ball is white and the second ball is blue,if the first ball drawn is not replaced in the bag?

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