If the difference between the mean and varian | vedclass

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(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

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