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)

Options

A. `(77)/(96)`
B. `(14)/(15)`
C. `(2)/(5)`
D. `(13)/(14)`

Answer

D

Explanation

`(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)`