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There are 5 bags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white balls is drawn at random. Find the probability that this white ball is from a bag of the first group. |
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Answer» <P> SOLUTION :Let `E_1` = event of selecting a bag from the first group,`E_2` = event of selecting a bag from the second group, and E = event of drawing a WHITE ball. Then, `P(E_1)=5/11and P(E_2)=6/11`. `P(E//E_1)`= probability of GETTING a white ball, given that it is from a bag of the first group `=5/8`. `P(E//E_2)` = probability of getting a white ball, given that it is from a bag of the second group `=2/6=1/3` Probability of getting the ball from a bag of the first group, given that it is white `P(E_1//E)` `=(P(E//E_1).P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2))`[by Bayes's theorem] `=((5/8xx5/11))/((5/8xx5/11)+(1/3xx6/11))=75/123`. |
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