In a hurdle race,a player has to cross 10 hur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $p$ be the probability of clearing a hurdle and $q$ be the probability of knocking down a hurdle.Given $p = \frac{5}{6}$,so $q = 1 - p = 1 - \

Vedclass · Accepted Answer

$\frac{5^{10}}{2 	imes 6^9}$

Vedclass · Answer

(B) Let $p$ be the probability of clearing a hurdle and $q$ be the probability of knocking down a hurdle.Given $p = \frac{5}{6}$,so $q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6}$.Let $X$ be the number of hurdles knocked down. Since there are $n = 10$ hurdles,$X$ follows a binomial distribution $B(n, q)$ where $n = 10$ and $q = \frac{1}{6}$ is the probability of success (knocking down a hurdle).The probability of knocking down fewer than $2$ hurdles is $P(X $P(X = x) = ^{10}C_x q^x p^{10-x}$.$P(X = 0) = ^{10}C_0 (\frac{1}{6})^0 (\frac{5}{6})^{10} = (\frac{5}{6})^{10}$.$P(X = 1) = ^{10}C_1 (\frac{1}{6})^1 (\frac{5}{6})^9 = 10 	imes \frac{1}{6} 	imes (\frac{5}{6})^9 = \frac{10}{6} 	imes (\frac{5}{6})^9 = \frac{5}{3} 	imes (\frac{5}{6})^9$.$P(X < 2) = (\frac{5}{6})^{10} + \frac{5}{3} 	imes (\frac{5}{6})^9 = (\frac{5}{6})^9 [\frac{5}{6} + \frac{5}{3}] = (\frac{5}{6})^9 [\frac{5+10}{6}] = (\frac{5}{6})^9 [\frac{15}{6}] = (\frac{5}{6})^9 [\frac{5}{2}] = \frac{5^{10}}{2 	imes 6^9}$.

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?

Similar Questions

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively,then $P(X < 3)$ equals to

Two coins are tossed $5$ times. What is the probability of getting an odd number of heads?

What is the probability of getting at least $1$ tail when a coin is tossed $4$ times?

If a dice is thrown twice,the probability of occurrence of $4$ at least once is

$A$ radar system can detect an enemy plane in one out of ten consecutive scans. The probability that it can detect an enemy plane at least twice in four consecutive scans is

Vedclass Test Series

Exam Paper Generator

Online Exam Module