**A bag containing a white ball and two red balls. A ball is drawn at random. If the ball is white which makes b**

*A bag containing a white ball and two* red balls. A ball is drawn at random. If the ball is white that is put back into the bag along with other white ball. If the ball is red that is returned held together with two extra balls red. A) Determine the probability that the second ball drawn is red. b) If the second ball drawn is red what is the probability that the first ball drawn is red?

Let A be the event that the first ball is white, 'first is the second B, red is red, the second B' is white. If A, then the second draw is 2 white and 2 red, and so P (B | A) = 2 / 4 = 1 / 2, and of course P (B | A) = 1 / 2 where P (B | A) is the probability of B given A has occurred. If B has 4 red and 1 white for the second draw, and thus (B | PA ') = 1 / 5 and P (B | A') = 4 / 5 The second ball is red if a followed by "B" or "A" occurs followed by 'B'. Now P (AB) = P (A) * P (B | A) ………………. = 1 / 3 * 1 / 2 ……………….. = 1 / 6 P (A'B ') = P (A') * P (B | A ') ………….= 2 / 3 * 4 / 5 … ……… = 8 / 15 Therefore, P (B ') = 1 / 6 + 8 / 15 ……………………. = 5 / 30 + 16/30 ……………………. = 21/30 ……………………. = 7 / 10 b) From above, looking at 30 equally likely outcomes, B ', occurs in 21 of these results. Of these 21, there are 5 in which A and 16, when a '[is say, the first red ball] occurs. Therefore problems first was red = 16/21. Formally, P (A'B ') = P (A' | B ') * P (B') We have seen that P (A'B ') = 8 / 15 and P (B) = 7 / 10 Consequently 8 / 15 = P (A' | B ') * 7 / 10 P (AB' | ') = 8 / 15 * 10 / 7 = 16/21

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