In a hurdles race- a player has to cross 10 hurdles- The probability that he will clear each hurdle is 56- What is the probability that he will knock down fewer than 2 hurdles -

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Standard XII

Mathematics

Independent Events

In a hurdles ...

Question

In a hurdles 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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Solution

Let X denote the number of hurdles knocked down by the player. Then, X follows binomial distribution with $n = 10$
$p = 1 - \frac{5}{6} = \frac{1}{6}$ and $q = \frac{5}{6}$
Therefore,
P(X = r) = ${10_{C}}_{3} (\frac{1}{6})^{r} (\frac{5}{6})^{10 - r}$ , $r = 0, 1, 2, . . ., 10$
Required probability
$= P (X < 2)$
$= P (X = 0) + P (X = 1)$
$= (\frac{5}{6})^{10} + {10_{C}}_{1} \times \frac{1}{6} \times (\frac{5}{6})^{9}$

$= (\frac{5}{6})^{10} [\frac{5}{6} + \frac{10}{6}] = \frac{5^{10}}{2 \times 6^{9}}$

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In a hurdles race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 56. What is the probability that he will knock down fewer than 2 hurdles ?

Let X denote the number of hurdles knocked down by the player. Then, X follows binomial distribution with n=10p=1−56=16 and q=56Therefore,P(X = r) = 10C3(16)r(56)10−r, r=0,1,2,...,10Required probability =P(X<2)=P(X=0)+P(X=1)=(56)10+10C1×16×(56)9=(56)10[56+106]=5102×69

In a hurdles 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 ?