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 $n = 6$. The probability mass function of a binomial distribution is

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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 $n = 6$. The probability mass function of a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$, where $q = 1-p$. According to the given condition, $P(X=2) = 4 P(X=4)$. Substituting the values: ${ }^6 C_2 p^2 q^4 = 4 \cdot { }^6 C_4 p^4 q^2$ Since ${ }^6 C_2 = 15$ and ${ }^6 C_4 = 15$, we have: $15 p^2 q^4 = 4 \cdot 15 p^4 q^2$ Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$): $q^2 = 4 p^2$ $(1-p)^2 = 4 p^2$ $1 - 2p + p^2 = 4 p^2$ $3p^2 + 2p - 1 = 0$ Factoring the quadratic equation: $(3p - 1)(p + 1) = 0$ This gives $p = \frac{1}{3}$ or $p = -1$. Since the probability $p$ must be in the interval $[0, 1]$, we discard $p = -1$. Therefore, $p = \frac{1}{3}$.

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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