If X is a binomial variate with the range {0, | vedclass

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(A) Given that $X$ is a binomial variate with range $\{0, 1, 2, 3, 4, 5, 6\}$,so the number of trials $n = 6$. The probability mass function of a bino

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$\frac{1}{3}$

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(A) Given that $X$ is a binomial variate with range $\{0, 1, 2, 3, 4, 5, 6\}$,so the number of trials $n = 6$. The probability mass function of a binomial distribution is given by $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$. According to the problem,$P(X=2) = 4 P(X=4)$. Substituting the values: ${^6C_2} p^2 q^4 = 4 \cdot {^6C_4} p^4 q^2$. Since ${^6C_2} = \frac{6 	imes 5}{2 	imes 1} = 15$ and ${^6C_4} = {^6C_2} = 15$,we have: $15 p^2 q^4 = 4 \cdot 15 p^4 q^2$. Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$): $q^2 = 4 p^2$. Substituting $q = 1-p$: $(1-p)^2 = 4 p^2$. Taking the square root on both sides: $1-p = 2p$ or $1-p = -2p$. Case $1$: $1 = 3p \Rightarrow p = \frac{1}{3}$. Case $2$: $1 = -p \Rightarrow p = -1$ (which is impossible as $0 \leq p \leq 1$). Therefore,the parameter $p = \frac{1}{3}$.

If $X$ is a binomial variate with the range $\{0, 1, 2, 3, 4, 5, 6\}$ and $P(X=2) = 4 P(X=4)$,then the parameter $p$ of $X$ is

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