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

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$\frac{1}{4}$

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(C) The probability mass function of a binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$,where $q = 1 - p$.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 given condition,$9P(X = 4) = P(X = 2)$.Substituting the values,we get $9 	imes {}^6C_4 p^4 q^2 = {}^6C_2 p^2 q^4$.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 $9 	imes 15 p^4 q^2 = 15 p^2 q^4$.Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$),we get $9p^2 = q^2$.Taking the square root of both sides,$3p = q$.Since $q = 1 - p$,we have $3p = 1 - p$,which implies $4p = 1$.Therefore,$p = \frac{1}{4}$.

If $X$ follows a binomial distribution with parameters $n = 6$ and $p$. If $9P(X = 4) = P(X = 2)$,then $p = $

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