# In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6 . What is the probability that he will knock down fewer than 2 hurdles?

This question has appeared in modelpaper 2012

Toolbox:
• A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
P (clearing a hurdle) $p = \large\frac{5}{6} \rightarrow$$q = 1- p = \large\frac{1}{6} This is a case of Bernoulli trials w n = 10 P(X=r)=C^{10}_{r} \large\frac{5}{6}^{10-r} \frac{1}{6}^{r} We need to calculate P (X \lt 2) = P (X = 0) + P (X = 1) P(X=0)=C^{10}_{0} \large\frac{5}{6}^{10} \frac{1}{6}^{0} = (\frac{5}{6})^{10} P(X=1)=C^{10}_{1} \large\frac{5}{6}^{10-1} \frac{1}{6}^{1} =$$ 10 \large(\frac{5}{6})^9 \frac{1}{6}$
$P (X \lt 2) =\large (\frac{5}{6})^{10}$$+ 10 \large(\frac{5}{6})^9 \frac{1}{6}$ $= \large(\frac{5}{6})^9 (\frac{5+10}{6})$
$= \large(\frac{5}{6})^9 (\frac{5}{2}) = \frac{5^{10}}{2\times6^9}$
Hey could you explain the last second step?
P(x<2) the last second step. How we reached the final answer