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(D) The probability mass function for a binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$,where $q = 1 - p$.Given $n = 6$,we have $P(

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$\frac{1}{3}$

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(D) The probability mass function for a binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$,where $q = 1 - p$.Given $n = 6$,we have $P(X = 4) = {}^6C_4 p^4 q^2$ and $P(X = 2) = {}^6C_2 p^2 q^4$.According to the problem,$4(P(X = 4)) = P(X = 2)$.Substituting the values,we get $4 	imes {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$.Since ${}^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ and ${}^6C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$,the equation becomes $4 	imes 15 p^4 q^2 = 15 p^2 q^4$.Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$),we get $4 p^2 = q^2$.Substituting $q = 1 - p$,we have $4 p^2 = (1 - p)^2$.$4 p^2 = 1 - 2p + p^2$.$3 p^2 + 2p - 1 = 0$.Factoring the quadratic equation: $(3p - 1)(p + 1) = 0$.Since $p$ must be a probability,$0 \le p \le 1$,so $p = \frac{1}{3}$.

If $X$ follows a binomial distribution with parameters $n = 6$ and $p$ and $4(P(X = 4)) = P(X = 2)$,then $p = $

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