Question

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

Answer 1

Answer: (A)

3/5

Answer 2

Let A denote the event that the runner succeeds exactly 3 times out of five and B denote the event that the runner succeeds on the first trial. P(B/A) = $\frac { \mathrm { P } ( \mathrm { B } \cap \mathrm { A } ) } { \mathrm { P } ( \mathrm { A } ) }$

But P (B Ē A) = P (clearing succeeding in the first trial and exactly once in two other trials)

P (⁴C₂ p² (1 – p)² ) = 6p³ (1 – p)²

and P (1) = ⁵C₃ p³ (1 – p)² = 10 p³ (1 – p)²

Thus, P(B/A) = $\frac { 6 p ^ { 3 } ( 1 - p ) ^ { 2 } } { 10 p ^ { 3 } ( 1 - p ) ^ { 2 } }$= $\frac { 3 } { 5 }$ .