Question

In $n$ independent trials (finite) of a random experiment, let $X$ be the number of times an event $A$ occurs. If the probability of success of one trial say $p$ and we get the probability of failure of event $A$ i.e. $P\left(\overline{A}\right)=1-p=q$. The probability of $r$ success in $n$ trials is denoted by$P(X=r)$ such that $P(X=r){=}^n{C}_r{p}^r{q}^{n-r}$, known as binomial distribution of random variable $X$. We also have the following result-

(i) Probability of getting at least $k$ successes is

$P(X\ge K)={\sum }_{r=k}^n{}^n{C}_r{p}^r{q}^{n-r}$

(ii) Probability of getting at the most $k$ successes is

$P(X\le K)={\sum }_{r=0}^n{}^n{C}_r{p}^r{q}^{n-r}$

(iii) ${\sum }_{r=0}^n{=}^n{C}_r{p}^r{q}^{n-r}={(p+q)}^n=1$

On the basis of above information answer the following question.

In a hurdle race, a player has to cross $10$ hurdles.The probability that he will cross each hurdle is $\frac{5}{6}$, then the probability that he will knock down fewer than two hurdle is

• A. $7{\left(\frac{5}{6}\right)}^{10}$
• B. $5{\left(\frac{5}{6}\right)}^{10}$
• C. $3{\left(\frac{5}{6}\right)}^{10}$
• D. $\frac{3}{5}{\left(\frac{5}{6}\right)}^{10}$

Answer 1

Answer: (C)

$3{\left(\frac{5}{6}\right)}^{10}$

Here ,$n=10,p=$ knocking down the hurdle $=\frac{1}{6}$

and $q=$ Clearing the hurdle $=\frac{5}{6}$

Using $P(X=r)={}^{10}{C}_r{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{10-r}$

$\therefore$ Required probability

$=P(X=r<2)=P\left(0\right)+P\left(1\right)={\left(\frac{5}{6}\right)}^{10}+\left(10\right)\frac{1}{6}{\left(\frac{5}{6}\right)}^9=(\frac{5}{6}+\frac{10}{6}){\left(\frac{5}{6}\right)}^9=3{\left(\frac{5}{6}\right)}^{10}$