A bag contains 7 black and 5 white balls. A b | vedclass

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(D) Total number of balls in the bag $= 7 + 5 = 12$.Since the ball is replaced after the first draw,the total number of balls remains $12$ for the sec

Vedclass · Accepted Answer

$\frac{35}{144}$

Vedclass · Answer

(D) Total number of balls in the bag $= 7 + 5 = 12$.Since the ball is replaced after the first draw,the total number of balls remains $12$ for the second draw.The probability of drawing a white ball in the first draw is $P(W) = \frac{5}{12}$.The probability of drawing a black ball in the second draw is $P(B) = \frac{7}{12}$.Since these are independent events,the probability that the first ball is white and the second ball is black is $P(W \cap B) = P(W) 	imes P(B) = \frac{5}{12} 	imes \frac{7}{12} = \frac{35}{144}$.

$Face$	$1, 2, 3, 4, 5, 6$
$Probability$	$0.1, 0.24, 0.19, 0.18, 0.15, 0.14$

$A$ bag contains $7$ black and $5$ white balls. $A$ ball is drawn out and replaced in the bag. Then a ball is drawn again. What is the probability that the first ball was white and the second black?

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