Step 1:

Let $X$ be the random variable denoting the number of hurdles out of 10 that a player knocks down in a race.

Probability that he will cross a hurdle =$\large\frac{5}{6}$$(q)$

$\therefore$ probability that he knock down a hurdle =$p=1-\large\frac{5}{6}=\frac{1}{6}$

$X\sim B(10,\large\frac{1}{6})$

Step 2:

The probability distribution of $X$ is given by

$P(X=x)=10C_x(\large\frac{1}{6})^x(\large\frac{5}{6})^{10-x}$$\qquad x=0,1,2........10$

Step 3:

Probability that the player will knock down less than two hurdles

$P(X < 2)=P(X=0)+P(X=1)$

$\qquad\qquad=10C_0(\large\frac{1}{6}(\large\frac{5}{6})^{10}$$+10C_1(\large\frac{1}{6})(\large\frac{5}{6})^9$

$\qquad\qquad=\large\frac{5^9}{6^{10}}$$[5+10]$

$\qquad\qquad=\large\frac{5^9}{6^{10}}$$\times 15$

$\qquad\qquad=\large\frac{15(5^9)}{6^{10}}$