Question

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear hurdles is $\frac{5}{6}$. What is the probability that he will knock down fewer than 2 hurdles?

Answer 1

It is a case of Bernoulli trials, where success is crossing a hurdle successfully without knocking it down and n=10.

$therefore$ p=P(success)$=\frac{5}{6}$

$\Rightarrow$ $q=1-p=\frac{1}{6}$.

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Clearly , X has a binomial distribution with $n=10andp=\frac{5}{6}$

$\therefore P(X=r){=}^n{C}_r{q}^{n-r}.p^r$

${=}^{10}{C}^r{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{10-r}$

P (player knocking down less than 2 hurdles) =$P(x\le 2)$

$=P\left(0\right)+P\left(1\right){=}^{10}{C}_0{p}^{10}{q}^0{+}^{10}{C}_1{p}^{9}q$

${=}^{1}0C_0{\left(\frac{5}{6}\right)}^{10}{\left(\frac{1}{6}\right)}^0{+}^{1}0C_1{\left(\frac{5}{6}\right)}^9{\left(\frac{1}{6}\right)}^1=1\times 1\times {\left(\frac{5}{6}\right)}^{10}+10\times {\left(\frac{5}{6}\right)}^9\times \left(\frac{1}{6}\right)={\left(\frac{5}{6}\right)}^9(\frac{5}{6}+\frac{10}{6})=\frac{15}{6}\times {\left(\frac{5}{6}\right)}^9=\frac{5}{2}\times \frac{5^9}{6^9}=\frac{5^{10}}{2\times 6^9}$