In a Binomial distribution with n=4,if 2 P(X= | vedclass

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(B) The probability mass function for a Binomial distribution is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.Given $n=4$,we have:$P(X=3) = { }^4 C_3

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$\frac{144}{169}$

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(B) The probability mass function for a Binomial distribution is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.Given $n=4$,we have:$P(X=3) = { }^4 C_3 p^3 (1-p)^1 = 4p^3(1-p)$$P(X=2) = { }^4 C_2 p^2 (1-p)^2 = 6p^2(1-p)^2$Given the condition $2 P(X=3) = 3 P(X=2)$:$2 	imes [4p^3(1-p)] = 3 	imes [6p^2(1-p)^2]$$8p^3(1-p) = 18p^2(1-p)^2$Dividing both sides by $2p^2(1-p)$ (assuming $p 
eq 0, 1$):$4p = 9(1-p)$$4p = 9 - 9p$$13p = 9 \implies p = \frac{9}{13}$Then $q = 1 - p = 1 - \frac{9}{13} = \frac{4}{13}$.The variance of a Binomial distribution is $npq$:$	ext{Variance} = 4 	imes \frac{9}{13} 	imes \frac{4}{13} = \frac{144}{169}$.

In a Binomial distribution with $n=4$,if $2 P(X=3)=3 P(X=2)$,then the variance is

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