Question

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

• A. $\frac{5^9}{2\times 6^9}$
• B. $\frac{5^{10}}{2\times 6^{10}}$
• C. $\frac{5^9}{2\times 6^{10}}$
• D. $\frac{5^{10}}{2\times 6^9}$

Answer 1

Answer: (D)

It is a case of Bernoulli trials, where success is not crossing a hurdle successfully. Here, $n=10$.
$p=P$ ( success ) $=1-\frac{5}{6}=\frac{1}{6}$
$\Rightarrow q=\frac{5}{6}$
let $X$ he the random variable that represents the numher of times the player will knock down the hurdle.
Clearly, $X$ has a binomial distribution with $n=10$ and $p$
$=\frac{1}{6}$
$\therefore P(X=r)={}^n{C}_r{q}^{n-r}\cdot p^r=10_{C_r}{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{10-r}$
P (player knocking down less than 2 hurdles)
$=P(X<2)=P(X=0)+P(X=1)$
$={}^{10}{C}_0{\left(\frac{1}{6}\right)}^0{\left(\frac{5}{6}\right)}^{10-0}+{}^{10}{C}_1{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^9$
$-{\left(\frac{5}{6}\right)}^9\left(\frac{5}{6}+\frac{10}{6}\right)=\frac{5^{10}}{2\times 6^9}$