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In a hurdle race a player has to cross $10$ hurdles. The probability that he will clear each hurdle is $\large\frac{5}{6}$. What is the probability that he will knock down less than $2$ hurdles.

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Toolbox:
  • A random variable $X$ is said to follow a binomial distribution of its probability mass function is given by
  • $P(X=x)=p(x)=\left\{\begin{array}{1 1}nC_xp^xq^{n-x},&x=0,1.......n\\0,&otherwise\end{array}\right.$
  • Constants of Binomial Distribution :
  • Mean = np
  • Variance = npq
  • Standard deviation =$\sqrt{ variance }=\sqrt{ npq}$
  • In a bionomial distribution Mean > Variance.
  • The parameters of the distribution are $n,p\quad X\sim B(n,p)$
Step 1:
Let $X$ be the random variable denoting the number of hurdles out of 10 that a player knocks down in a race.
Probability that he will cross a hurdle =$\large\frac{5}{6}$$(q)$
$\therefore$ probability that he knock down a hurdle =$p=1-\large\frac{5}{6}=\frac{1}{6}$
$X\sim B(10,\large\frac{1}{6})$
Step 2:
The probability distribution of $X$ is given by
$P(X=x)=10C_x(\large\frac{1}{6})^x(\large\frac{5}{6})^{10-x}$$\qquad x=0,1,2........10$
Step 3:
Probability that the player will knock down less than two hurdles
$P(X < 2)=P(X=0)+P(X=1)$
$\qquad\qquad=10C_0(\large\frac{1}{6}(\large\frac{5}{6})^{10}$$+10C_1(\large\frac{1}{6})(\large\frac{5}{6})^9$
$\qquad\qquad=\large\frac{5^9}{6^{10}}$$[5+10]$
$\qquad\qquad=\large\frac{5^9}{6^{10}}$$\times 15$
$\qquad\qquad=\large\frac{15(5^9)}{6^{10}}$
answered Sep 18, 2013 by sreemathi.v
 

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