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) $\dfrac{3}{5}$
B) $\dfrac{2}{5}$
C) $\dfrac{1}{5}$
D) $\dfrac{4}{5}$

Answer 1

Find the probability of not jumping over a hurdle is 1 minus probability of jumping over a hurdle, then use the concept of combination to find the success 3 trials out of 5.

Complete step by step answer :
As given in the question, the probability of jumping over a specific hurdle is$p$, therefore the probability of not jumping will be $1 - p$.
According to the question, he was able to succeed in 3 trials out of 5 trials. Probability of this situation will be $\mathop C\nolimits_3^5 {\left( p \right)^3}{(1 - p)^2}$.
Probability of runner being successful in the first attempt will be $p\mathop C\nolimits_2^4 {\left( p \right)^2}{(1 - p)^2}$, where first $p$being the probability of jumping over the first hurdle , and then there will be 4 trials left in which he will succeed in jumping over 2 out of 4 hurdles, that is why $\mathop C\nolimits_2^4$is mentioned.${p^2}$is the probability of successfully jumping over the hurdles twice, while ${(1 - p)^2}$is for not being able to jump over the remaining hurdles.
Required probability of a runner succeeding in jumping over the first hurdle is given by the probability of jumping over the first hurdle divided by the probability of jumping over the hurdles successfully 3 times in 5 trials.
Required probability=$\dfrac{{\mathop C\nolimits_2^4 {{\left( p \right)}^3}{{(1 - p)}^2}}}{{\mathop C\nolimits_3^5 {{\left( p \right)}^3}{{(1 - p)}^2}}} = \dfrac{{\mathop C\nolimits_2^4 {{\left( p \right)}^{3 - 3}}{{(1 - p)}^{2 - 2}}}}{{\mathop C\nolimits_3^5 }}$
$\dfrac{{\mathop C\nolimits_2^4 }}{{\mathop C\nolimits_3^5 }} = \dfrac{{\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{5!}}{{3!\left( {5 - 3} \right)!}}}} \\\ = \dfrac{{\dfrac{{4!}}{{2!\left( 2 \right)!}}}}{{\dfrac{{5!}}{{3!\left( 2 \right)!}}}} \\\ = \dfrac{{4!}}{{2!\left( 2 \right)!}} \cdot \dfrac{{3!\left( 2 \right)!}}{{5!}} \\\ = \dfrac{{4 \cdot 3 \cdot 2 \cdot 1}}{{2 \cdot 1 \cdot 2 \cdot 1}} \cdot \dfrac{{3 \cdot 2 \cdot 1 \cdot 2 \cdot 1}}{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}} \\\ = \dfrac{3}{5} \\\$
so, the probability of not succeeding is $\dfrac{3}{5}$
Hence, the answer is option A which is $\dfrac{3}{5}$.

Note: Probability of succeeding and not succeeding should be taken into consideration to arrive at the answer.