InterviewSolution
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InterviewSolution
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Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3,4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let `x_i` be the number on the card drawn from the ith box, i = 1, 2, 3.The probability that `x_1+x_2+x_3` is odd isThe probability that `x_1, x_2, x_3` are in an aritmetic progression isA. `(29)/(105)`B. `(53)/(105)`C. `(57)/(105)`D. `(1)/(2)` |
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Answer» Correct Answer - B Fo, `x_(1) + x_(2) + x_(3)` = odd. Case I: One odd, two even (OEE) or (EOE) or (EEO) Total number of ways = `2xx2xx3+1xx3xx3+1xx2xx4=29`. Case II: All three odd Number of ways = `2xx3xx4=24` `therefore` Favorable ways = 53 `therefore` Required probability = `(53)/(3xx5xx7) = (53)/(105)` |
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