For a binomial variate X with n=6,if P(X=2)=9 | vedclass

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(C) Given that $n=6$ and $P(X=2)=9 P(X=4)$.Using the binomial probability formula $P(X=k) = {}^n C_k p^k q^{n-k}$:${}^6 C_2 p^2 q^4 = 9 \cdot {}^6 C_4

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$\frac{9}{8}$

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(C) Given that $n=6$ and $P(X=2)=9 P(X=4)$.Using the binomial probability formula $P(X=k) = {}^n C_k p^k q^{n-k}$:${}^6 C_2 p^2 q^4 = 9 \cdot {}^6 C_4 p^4 q^2$Since ${}^6 C_2 = 15$ and ${}^6 C_4 = 15$,we have:$15 p^2 q^4 = 9 \cdot 15 p^4 q^2$Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$):$q^2 = 9 p^2$Taking the square root on both sides:$q = 3p$Since $p + q = 1$,we substitute $q = 3p$:$p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$Then $q = 1 - \frac{1}{4} = \frac{3}{4}$.The variance of a binomial distribution is given by $npq$:$	ext{Variance} = 6 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

For a binomial variate $X$ with $n=6$,if $P(X=2)=9 P(X=4)$,then its variance is

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