**Question:**

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

he will knock down fewer than 2 hurdles?

**Solution:**

Let *p* and *q* respectively be the probabilities that the player will clear and knock down the hurdle.

$\therefore p=\frac{5}{6}$

$\Rightarrow q=1-p=1-\frac{5}{6}=\frac{1}{6}$

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

Therefore, by binomial distribution, we obtain

$\mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} p^{n-x} q^{x}$

P (player knocking down less than 2 hurdles) = P (X < 2)

= P (X = 0) + P (X = 1)

$={ }^{10} C_{0}(q)^{0}(p)^{10}+{ }^{10} C_{1}(q)(p)^{9}$

$=\left(\frac{5}{6}\right)^{10}+10 \cdot \frac{1}{6} \cdot\left(\frac{5}{6}\right)^{9}$

$=\left(\frac{5}{6}\right)^{9}\left[\frac{5}{6}+\frac{10}{6}\right]$

$=\frac{5}{2}\left(\frac{5}{6}\right)^{9}$

$=\frac{(5)^{10}}{2 \times(6)^{9}}$