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(C) The probability mass function of a Binomial distribution $B(n, p)$ is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n=10$,we have $P(X=k

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$\frac{12}{5}$

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(C) The probability mass function of a Binomial distribution $B(n, p)$ is given by $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.Given $n=10$,we have $P(X=k) = \binom{10}{k} p^k (1-p)^{10-k}$.Given $5 P(X=0) = P(X=1)$:$5 \binom{10}{0} p^0 (1-p)^{10} = \binom{10}{1} p^1 (1-p)^9$$5(1-p) = 10p$$5 - 5p = 10p \implies 15p = 5 \implies p = \frac{1}{3}$.Thus,$q = 1-p = \frac{2}{3}$.Now,we calculate the ratio $\frac{P(X=5)}{P(X=6)}$:$\frac{P(X=5)}{P(X=6)} = \frac{\binom{10}{5} p^5 q^5}{\binom{10}{6} p^6 q^4} = \frac{\binom{10}{5}}{\binom{10}{6}} \cdot \frac{q}{p}$$\binom{10}{5} = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 252$$\binom{10}{6} = \binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$\frac{P(X=5)}{P(X=6)} = \frac{252}{210} \cdot \frac{2/3}{1/3} = \frac{6}{5} \cdot 2 = \frac{12}{5}$.

If a random variable $X$ follows the Binomial distribution $B(10, p)$ such that $5 P(X=0) = P(X=1)$,then the value of $\frac{P(X=5)}{P(X=6)}$ is equal to

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