In a hurdle race,a player has to cross 10 hur | vedclass

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(B) Let $p$ be the probability of clearing a hurdle and $q$ be the probability of knocking down a hurdle.Given $p = \frac{5}{6}$,so $q = 1 - p = 1 - \

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$\frac{5^{10}}{2 	imes 6^9}$

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(B) Let $p$ be the probability of clearing a hurdle and $q$ be the probability of knocking down a hurdle.Given $p = \frac{5}{6}$,so $q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6}$.Let $X$ be the number of hurdles knocked down. Since there are $n = 10$ hurdles,$X$ follows a binomial distribution $B(n, q)$ where $n = 10$ and $q = \frac{1}{6}$ is the probability of success (knocking down a hurdle).The probability of knocking down fewer than $2$ hurdles is $P(X $P(X = x) = ^{10}C_x q^x p^{10-x}$.$P(X = 0) = ^{10}C_0 (\frac{1}{6})^0 (\frac{5}{6})^{10} = (\frac{5}{6})^{10}$.$P(X = 1) = ^{10}C_1 (\frac{1}{6})^1 (\frac{5}{6})^9 = 10 	imes \frac{1}{6} 	imes (\frac{5}{6})^9 = \frac{10}{6} 	imes (\frac{5}{6})^9 = \frac{5}{3} 	imes (\frac{5}{6})^9$.$P(X < 2) = (\frac{5}{6})^{10} + \frac{5}{3} 	imes (\frac{5}{6})^9 = (\frac{5}{6})^9 [\frac{5}{6} + \frac{5}{3}] = (\frac{5}{6})^9 [\frac{5+10}{6}] = (\frac{5}{6})^9 [\frac{15}{6}] = (\frac{5}{6})^9 [\frac{5}{2}] = \frac{5^{10}}{2 	imes 6^9}$.

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?

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