Two balls are drawn at random with replacemen | vedclass

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(A) Total number of balls $= 10 + 8 = 18$. Probability of drawing a black ball in the first draw,$P(B_1) = \frac{10}{18} = \frac{5}{9}$. Since the bal

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$\frac{20}{81}$

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(A) Total number of balls $= 10 + 8 = 18$. Probability of drawing a black ball in the first draw,$P(B_1) = \frac{10}{18} = \frac{5}{9}$. Since the ball is replaced,the total number of balls remains $18$ for the second draw. Probability of drawing a red ball in the second draw,$P(R_2) = \frac{8}{18} = \frac{4}{9}$. Since the events are independent,the probability that the first ball is black and the second is red is $P(B_1 \cap R_2) = P(B_1) 	imes P(R_2) = \frac{5}{9} 	imes \frac{4}{9} = \frac{20}{81}$.

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that the first ball is black and the second is red.

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