A binomial random variable X satisfies 9 cdot | vedclass

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Question

(A) 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)

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$\frac{1}{4}$

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(A) 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 $P(X=2) = {^6C_2} p^2 q^4$. The condition is $9 \cdot P(X=4) = P(X=2)$. Substituting the values: $9 \cdot {^6C_4} p^4 q^2 = {^6C_2} p^2 q^4$. 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: $9 \cdot 15 \cdot p^4 q^2 = 15 \cdot p^2 q^4$. Dividing both sides by $15 p^2 q^2$ (assuming $p, q 
eq 0$): $9 p^2 = q^2$. Taking the square root of both sides: $3p = q$. Since $q = 1-p$,we have $3p = 1-p$. $4p = 1$,which gives $p = \frac{1}{4}$.

$A$ binomial random variable $X$ satisfies $9 \cdot P(X=4) = P(X=2)$ when $n=6$. Then $p$ is equal to

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