A bag contains 4 red and 6 black balls. A bal | vedclass

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(A) Let $R_1$ be the event of drawing a red ball in the first draw and $B_1$ be the event of drawing a black ball in the first draw.Let $R_2$ be the e

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$\frac{2}{5}$

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(A) Let $R_1$ be the event of drawing a red ball in the first draw and $B_1$ be the event of drawing a black ball in the first draw.Let $R_2$ be the event of drawing a red ball in the second draw.Initially,the bag contains $4$ red and $6$ black balls,total $10$ balls.$P(R_1) = \frac{4}{10} = \frac{2}{5}$ and $P(B_1) = \frac{6}{10} = \frac{3}{5}$.If a red ball is drawn first,it is returned along with $2$ additional red balls. The bag now contains $4 + 2 = 6$ red balls and $6$ black balls,total $12$ balls.So,$P(R_2 | R_1) = \frac{6}{12} = \frac{1}{2}$.If a black ball is drawn first,it is returned along with $2$ additional black balls. The bag now contains $4$ red balls and $6 + 2 = 8$ black balls,total $12$ balls.So,$P(R_2 | B_1) = \frac{4}{12} = \frac{1}{3}$.Using the law of total probability,the probability that the second ball is red is:$P(R_2) = P(R_1) 	imes P(R_2 | R_1) + P(B_1) 	imes P(R_2 | B_1)$$P(R_2) = (\frac{4}{10} 	imes \frac{6}{12}) + (\frac{6}{10} 	imes \frac{4}{12})$$P(R_2) = \frac{24}{120} + \frac{24}{120} = \frac{48}{120} = \frac{2}{5}$.

$A$ bag contains $4$ red and $6$ black balls. $A$ ball is drawn at random from the bag,its colour is observed,and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red,is:

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