InterviewSolution
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InterviewSolution
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A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag a is chosen, otherwise bag B. If two balls are drawn (without replacement) from the selected bag, find the probability of one of them being red and another black. |
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Answer» Given that, bag A contains 4 black (B) & 6 red (R) balls bag B contains 7 black (B) & 3 Red (R) balls let the probability of Bag A be P(A) and B be P(B) respectively. so, P(A) = `2/6 = 1/3` and P(B)= `1- 1/3 = 2/3` Now, probability of 1B,1R `P(1B,1R) = P(A)*P((1B,1R)/A) + P(B)*P((1B,1R)/B)` = `1/3*(.^4C_1 * .^6C_1)/(.^10C_2)`+ `2/3*(.^7C_1*.^3C_1)/(.^10C_2)` = `8/45 + 14/45 = 22/45` answer |
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