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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
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The correct answer is:
\(\frac{5^{10}}{2 \times 6^9}\)
It is a case of Bernoulli trials, where success is not crossing a hurdle successfully. Here, \(\mathrm{n}=\) 10.
\(\mathrm{p}=\mathrm{P}\text { (success) }=1-\frac{5}{6}=\frac{1}{6} \Rightarrow \mathrm{q}=\frac{5}{6}\)
let \(\mathrm{X}\) be the random variable that represents the number of times the player will knock down the hurdle.
Clearly, \(\mathrm{X}\) has a binomial distribution with \(\mathrm{n}=\)
\(\begin{aligned}
10 \text { and } \mathrm{p}= & \frac{1}{6} \\
\therefore \mathrm{P}(\mathrm{X}=\mathrm{r}) & ={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \\
& ={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{6}\right)^{\mathrm{r}}\left(\frac{5}{6}\right)^{10-\mathrm{r}}
\end{aligned}\)
\(\mathrm{P}\) (player knocking down less than 2 hurdles)
\(\begin{aligned}
& =\mathrm{P}(\mathrm{X} < 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\
& ={ }^{10} \mathrm{C}_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{10-0}+{ }^{10} \mathrm{C}_1\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^9 \\
& =\left(\frac{5}{6}\right)^9\left(\frac{5}{6}+\frac{10}{6}\right)=\frac{5^{10}}{2 \times 6^9}
\end{aligned}\)
\(\mathrm{p}=\mathrm{P}\text { (success) }=1-\frac{5}{6}=\frac{1}{6} \Rightarrow \mathrm{q}=\frac{5}{6}\)
let \(\mathrm{X}\) be the random variable that represents the number of times the player will knock down the hurdle.
Clearly, \(\mathrm{X}\) has a binomial distribution with \(\mathrm{n}=\)
\(\begin{aligned}
10 \text { and } \mathrm{p}= & \frac{1}{6} \\
\therefore \mathrm{P}(\mathrm{X}=\mathrm{r}) & ={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{p}^{\mathrm{r}} \\
& ={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{6}\right)^{\mathrm{r}}\left(\frac{5}{6}\right)^{10-\mathrm{r}}
\end{aligned}\)
\(\mathrm{P}\) (player knocking down less than 2 hurdles)
\(\begin{aligned}
& =\mathrm{P}(\mathrm{X} < 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\
& ={ }^{10} \mathrm{C}_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{10-0}+{ }^{10} \mathrm{C}_1\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^9 \\
& =\left(\frac{5}{6}\right)^9\left(\frac{5}{6}+\frac{10}{6}\right)=\frac{5^{10}}{2 \times 6^9}
\end{aligned}\)
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