If a random variable X follows the Binomial d | vedclass

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(C) For a binomial distribution $B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$ and $n = 33$.

Vedclass · Accepted Answer

$\frac{121}{48}$

Vedclass · Answer

(C) For a binomial distribution $B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$ and $n = 33$. Given $3 P(X=0) = P(X=1)$,we substitute the values: $3 \binom{33}{0} p^0 q^{33} = \binom{33}{1} p^1 q^{32}$ $3 	imes 1 	imes q^{33} = 33 	imes p 	imes q^{32}$ Since $q 
eq 0$,we divide both sides by $q^{32}$: $3q = 33p$ $q = 11p$ Since $q = 1-p$,we have $1-p = 11p$,which implies $1 = 12p$,so $p = \frac{1}{12}$. Then $q = 1 - \frac{1}{12} = \frac{11}{12}$. The variance of a binomial distribution is given by $Var(X) = npq$. $Var(X) = 33 	imes \frac{1}{12} 	imes \frac{11}{12} = \frac{33 	imes 11}{144} = \frac{363}{144}$. Simplifying by dividing by $3$: $\frac{121}{48}$.

If a random variable $X$ follows the Binomial distribution $B(33, p)$ such that $3 P(X=0) = P(X=1)$,then the variance of $X$ is

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