If X sim B(9, p) is a binomial variate satisf | vedclass

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(A) The probability mass function of a binomial distribution $X \sim B(n, p)$ is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n=9$,the equa

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$\frac{23}{256}$

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(A) The probability mass function of a binomial distribution $X \sim B(n, p)$ is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n=9$,the equation $P(X=3)=P(X=6)$ implies:$\binom{9}{3} p^3 (1-p)^{9-3} = \binom{9}{6} p^6 (1-p)^{9-6}$Since $\binom{9}{3} = \binom{9}{6}$,we have:$p^3 (1-p)^6 = p^6 (1-p)^3$Dividing both sides by $p^3 (1-p)^3$ (assuming $p 
eq 0, 1$):$(1-p)^3 = p^3$$1-p = p \implies 2p = 1 \implies p = \frac{1}{2}$.Now,we need to find $P(X $P(X=0) = \binom{9}{0} (\frac{1}{2})^0 (\frac{1}{2})^9 = 1 \cdot 1 \cdot \frac{1}{512} = \frac{1}{512}$.$P(X=1) = \binom{9}{1} (\frac{1}{2})^1 (\frac{1}{2})^8 = 9 \cdot \frac{1}{512} = \frac{9}{512}$.$P(X=2) = \binom{9}{2} (\frac{1}{2})^2 (\frac{1}{2})^7 = 36 \cdot \frac{1}{512} = \frac{36}{512}$.$P(X < 3) = \frac{1+9+36}{512} = \frac{46}{512} = \frac{23}{256}$.

If $X \sim B(9, p)$ is a binomial variate satisfying the equation $P(X=3)=P(X=6)$,then $P(X < 3)=$

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