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There are two bags I and II.Bag I contains 3 white and 3 red balls and Bag II contains 4 white and 5 redballs. One ball is drawn at random from one of the bags and is found to bered. Find the probability that it was drawn from bag II. |
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Answer» Let `P(B_1)` and `P(B_2)` are the proababilities of seleting bag 1 and bag 2. Let `P(R)` is the probability of drawing red ball. Then, required probability can be given as, `P(B_2/R) = (P(R/B_2)P(B_2))/(P(R/B_2)P(B_2)+P(R/B_1)P(B_1))` Here, `P(R/B_2) = 5/9` `P(B_1) = P(B_2) = 1/2` `P(R/B_1) =3/6 = 1/2` Putting all these values, `P(B_2/R) = (5/9*1/2)/(5/9*1/2+1/2*1/2) = (5/18)/(5/18+1/4)` `=5/18*36/19 = 10/19` So, the required probability is `10/19`. |
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