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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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