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In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is $\frac{5}{6}$. What is the probability that he will knock down fewer than 2 hurdles?
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Verified Answer
There are 10 hurdles in all i.e., $\mathrm{n}=10$ He clears a hurdle, the probability is $=\frac{5}{6}$
$\therefore \quad$ probability that he does not clear the hurdle $=1-\frac{5}{6}=\frac{1}{6}$
$P($ knocking down fewer than 2$)$
$=\mathrm{P}$ (be will cross all hurdles) $+\mathrm{P}$ (He will not cross 1 hurdle)
$$
=\left(\frac{5}{6}\right)^{10}+{ }^{10} \mathrm{C}_1\left(\frac{5}{6}\right)^9\left(\frac{1}{6}\right)
$$
$$
=\left(\frac{5}{6}\right)^9\left[\frac{5}{6}+10 \times \frac{1}{6}\right]=\left(\frac{5}{6}\right)^9 \times \frac{15}{6}=\frac{5}{2}\left(\frac{5}{6}\right)^9
$$
$\therefore \quad$ probability that he does not clear the hurdle $=1-\frac{5}{6}=\frac{1}{6}$
$P($ knocking down fewer than 2$)$
$=\mathrm{P}$ (be will cross all hurdles) $+\mathrm{P}$ (He will not cross 1 hurdle)
$$
=\left(\frac{5}{6}\right)^{10}+{ }^{10} \mathrm{C}_1\left(\frac{5}{6}\right)^9\left(\frac{1}{6}\right)
$$
$$
=\left(\frac{5}{6}\right)^9\left[\frac{5}{6}+10 \times \frac{1}{6}\right]=\left(\frac{5}{6}\right)^9 \times \frac{15}{6}=\frac{5}{2}\left(\frac{5}{6}\right)^9
$$
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