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

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

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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 $P(X=k) = {^nC_k} p^k q^{n-k}$:${^6C_2} p^2 q^4 = 9 \cdot {^6C_4} p^4 q^2$Since ${^6C_2} = 15$ and ${^6C_4} = 15$, we have:$15 p^2 q^4 = 9 \cdot 15 p^4 q^2$$q^2 = 9 p^2$$q = 3p$ (since $p, q > 0$).We know that $p+q=1$, so $p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$.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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