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Q. 6 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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Let p and q respectively be probability that the player will clear and knock down the hurdle.

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

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

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

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

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

$=^{10}C_0 .( \frac{5}{6})^{10}.( \frac{1}{6})^{0}+^{10}C_1 .( \frac{5}{6})^{9}.( \frac{1}{6})^{1}$

$=( \frac{5}{6})^{10}+10.( \frac{5}{6})^{9}.( \frac{1}{6})$

$=( \frac{5}{6})^{9}( \frac{5}{6}+10\times . \frac{1}{6})$

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

$=\frac{5^1^0}{2\times 6^9}$

Posted by

seema garhwal

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