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} $
$-\left(\frac{5}{6}\right)^{9}\left(\frac{5}{6}+\frac{10}{6}\right)=\frac{5^{10}}{2 \times 6^{9}}$