In a hurdle race, a runner has probability `p` of jumping over a specific hurdle. Given that in `5` trials, the runner succeeded `3` times, the conditional probability that the runner had succeeded in the first trial is
(a) `3//5` (b)`2//5` (c) `1//5` (d) None of these

Options

A. `3//5`
B. `2//5`
C. `1//5`
D. None of these

Answer

A

Explanation

`(a)` Let `A` denote the event that the runner succeeds exactly `3` times out of five and `B` denote the events that the runner suceeds on the first trial.
`P(B//A)=(P(BnnA))/(P(A))`
But `P(BnnA)=P` (clearing succeeding in the first trial and exactly once in two other trials)
`=^(4)C_(2)p^(2)(1-p)^(2))=6p^(3)(1-p)^(2)`
and `P(A)=^(5)C_(3)p^(3)(1-p)^(2)=10p^(3)(1-p)^(2)`
Thus, `P(B//A)=(6p^(3)(1-p)^(2))/(10p^(3)(1-p)^(2))=(3)/(5)`