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

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(C) Total number of balls $= 7 + 5 = 12$.The probability of drawing a black ball in the first draw is $P(B_1) = \frac{7}{12}$.Since the ball is replac

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$\frac{49}{144}$

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(C) Total number of balls $= 7 + 5 = 12$.The probability of drawing a black ball in the first draw is $P(B_1) = \frac{7}{12}$.Since the ball is replaced,the total number of balls remains $12$ for the second draw.The probability of drawing a black ball in the second draw is $P(B_2) = \frac{7}{12}$.Since the events are independent,the probability that both balls are black is $P(B_1 \cap B_2) = P(B_1) 	imes P(B_2) = \frac{7}{12} 	imes \frac{7}{12} = \frac{49}{144}$.

$A$ 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 both balls were black?

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