A random variable X sim B(n, p) follows a bin | vedclass

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(A) The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n = 6$, we have $P(X = 4) = \binom{6

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$1/4$

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(A) The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n = 6$, we have $P(X = 4) = \binom{6}{4} p^4 (1-p)^2 = 15 p^4 (1-p)^2$ and $P(X = 2) = \binom{6}{2} p^2 (1-p)^4 = 15 p^2 (1-p)^4$.Given $9P(X = 4) = P(X = 2)$, substitute the expressions:$9 	imes 15 p^4 (1-p)^2 = 15 p^2 (1-p)^4$.Divide both sides by $15 p^2 (1-p)^2$ (assuming $p 
eq 0, 1$):$9 p^2 = (1-p)^2$.Taking the square root of both sides:$3p = 1-p$ or $3p = -(1-p)$.For $3p = 1-p$, we get $4p = 1$, so $p = 1/4$.For $3p = p-1$, we get $2p = -1$, which gives $p = -1/2$ (not possible as $0 \le p \le 1$).Thus, $p = 1/4$.

$A$ random variable $X \sim B(n, p)$ follows a binomial distribution with $n = 6$. If $9P(X = 4) = P(X = 2)$, then the probability of success $p$ is:

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