If X sim B(7, p) is a binomial variate and P( | vedclass

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(C) Given $X \sim B(n, p)$ with $n=7$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $P(X=3) = P(X=5)$, we have:$\bin

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$\frac{5-\sqrt{15}}{2}$

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(C) Given $X \sim B(n, p)$ with $n=7$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $P(X=3) = P(X=5)$, we have:$\binom{7}{3} p^3 (1-p)^{7-3} = \binom{7}{5} p^5 (1-p)^{7-5}$$\binom{7}{3} p^3 (1-p)^4 = \binom{7}{5} p^5 (1-p)^2$Since $\binom{7}{3} = \binom{7}{4} = 35$ and $\binom{7}{5} = \binom{7}{2} = 21$, we get:$35 p^3 (1-p)^4 = 21 p^5 (1-p)^2$Dividing both sides by $7 p^3 (1-p)^2$ (assuming $p 
eq 0, 1$):$5 (1-p)^2 = 3 p^2$$5 (1 - 2p + p^2) = 3 p^2$$5 - 10p + 5p^2 = 3p^2$$2p^2 - 10p + 5 = 0$Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{10 \pm \sqrt{100 - 4(2)(5)}}{2(2)} = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2}$Since $0 \le p \le 1$, we take $p = \frac{5 - \sqrt{15}}{2}$.

If $X \sim B(7, p)$ is a binomial variate and $P(X=3)=P(X=5)$ then $p=$

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