An urn contains 5 red and 5 black balls. A ba | vedclass

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(A) Let $R_1$ be the event that the first ball drawn is red and $B_1$ be the event that the first ball drawn is black.$P(R_1) = 5/10 = 1/2$ and $P(B_1

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$1/2$

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(A) Let $R_1$ be the event that the first ball drawn is red and $B_1$ be the event that the first ball drawn is black.$P(R_1) = 5/10 = 1/2$ and $P(B_1) = 5/10 = 1/2$.If $R_1$ occurs,$2$ red balls are added. The urn now has $7$ red and $5$ black balls (total $12$). The probability of drawing a red ball second is $P(R_2|R_1) = 7/12$.If $B_1$ occurs,$2$ black balls are added. The urn now has $5$ red and $7$ black balls (total $12$). The probability of drawing a red ball second is $P(R_2|B_1) = 5/12$.By the law of total probability,$P(R_2) = P(R_1)P(R_2|R_1) + P(B_1)P(R_2|B_1)$.$P(R_2) = (1/2 	imes 7/12) + (1/2 	imes 5/12) = 7/24 + 5/24 = 12/24 = 1/2$.

An urn contains $5$ red and $5$ black balls. $A$ ball is drawn at random,its colour is noted and is returned to the urn. Moreover,$2$ additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

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