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 trail is

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (A)

$\frac{3}{5}$

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.

$P\left(B|A\right)=\frac{P(B\cap A)}{P\left(A\right)}$

But $P(B\cap A)=P($ succeeding in the first trial and exactly once in two other trials$)$

$=p({}^4{C}_2.p^2{(1-p)}^2)=6p^3{(1-p)}^2$

and $P\left(A\right){=}^5{C}_3{p}^3{(1-p)}^2=10p^3{(1-p)}^2$

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