InterviewSolution
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InterviewSolution
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The probabilities of solving a problem correctly by `A` and `B` are `(1)/(8)` and `(1)/(12)` respectively. Given that they obtain the same answer after solving a problem and the probability of a common mistake by them is `(1)/(1001)`, then probability that their solution is correct is (Assuming that if they commit different mistake, then their answers will differ)A. `(77)/(96)`B. `(14)/(15)`C. `(2)/(5)`D. `(13)/(14)` |
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Answer» Correct Answer - D `(d)` Probability that they both solve correctly is `(1)/(8)xx(1)/(12)=(1)/(96)` Probability that they both get wrong solution is `(7)/(8)xx(11)/(12)=(77)/(96)` If `E_(1)` be the event of both getting correct solution and `E_(2)` be the event that both gets wrong solution. Let `E` be even of both obtaining the same answer, then `P((E_(1))/(E_()))=(P(E_(1))P((E)/(E_(1))))/(P(E_(1))P((E)/(E_(1)))+P(E_(2))P((E)/(E_(2))))` `=((1)/(96)xx1)/((1)/(96)xx1+(77)/(96)xx(1)/(1001))=(13)/(14)` |
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