If the difference between the mean and varian | vedclass

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

Question

(A) For a binomial distribution,Mean $= np$ and Variance $= npq$,where $n=5$ is the number of trials,$p$ is the probability of success,and $q=1-p$ is

Vedclass · Accepted Answer

$\frac{80}{243}$

Vedclass · Answer

(A) For a binomial distribution,Mean $= np$ and Variance $= npq$,where $n=5$ is the number of trials,$p$ is the probability of success,and $q=1-p$ is the probability of failure. Given that the difference between mean and variance is $\frac{5}{9}$: $np - npq = \frac{5}{9}$ $np(1-q) = \frac{5}{9}$ Since $1-q = p$,we have $np^2 = \frac{5}{9}$. Substituting $n=5$: $5p^2 = \frac{5}{9} \Rightarrow p^2 = \frac{1}{9} \Rightarrow p = \frac{1}{3}$. Then $q = 1 - \frac{1}{3} = \frac{2}{3}$. The probability of getting $2$ successes in $5$ trials is given by the binomial formula $P(X=k) = {^nC_k} p^k q^{n-k}$: $P(X=2) = {^5C_2} 	imes (\frac{1}{3})^2 	imes (\frac{2}{3})^{5-2}$ $P(X=2) = 10 	imes \frac{1}{9} 	imes \frac{8}{27} = \frac{80}{243}$.

If the difference between the mean and variance of a binomial variate is $\frac{5}{9}$,then the probability of getting $2$ successes for an event when the experiment is conducted $5$ times,is

Similar Questions

Find the binomial probability distribution whose mean is $3$ and variance is $2$.

The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $P(X=1)=\frac{4}{243}$,then $P(X=4 \text{ or } 5)$ is equal to.

The variance of the following probability distribution of $X$ is given by:
$X: 0, 1, 2, 3$
$P(X): q^3, 3pq^2, 3p^2q, p^3$
where $0 < p < 1$ and $q = 1 - p$.

In a box containing $100$ bulbs,$10$ are defective. The probability that out of a sample of $5$ bulbs,none is defective is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module