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(B) Given that $X$ follows a binomial distribution $B(n, p)$ with $n = 7$.The probability mass function is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^

Vedclass · Accepted Answer

$\frac{77}{36}$

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(B) Given that $X$ follows a binomial distribution $B(n, p)$ with $n = 7$.The probability mass function is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$.Given the condition $P(X=3) = 5P(X=4)$:${}^{7}C_{3} p^{3} (1-p)^{4} = 5 	imes {}^{7}C_{4} p^{4} (1-p)^{3}$.Since ${}^{7}C_{3} = {}^{7}C_{4} = 35$,we can simplify the equation:$35 p^{3} (1-p)^{4} = 5 	imes 35 p^{4} (1-p)^{3}$.Dividing both sides by $35 p^{3} (1-p)^{3}$ (assuming $p 
eq 0, 1$):$(1-p) = 5p$.$1 = 6p \Rightarrow p = \frac{1}{6}$.Thus,$q = 1 - p = \frac{5}{6}$.Mean $= np = 7 	imes \frac{1}{6} = \frac{7}{6}$.Variance $= npq = 7 	imes \frac{1}{6} 	imes \frac{5}{6} = \frac{35}{36}$.Sum of mean and variance $= \frac{7}{6} + \frac{35}{36} = \frac{42 + 35}{36} = \frac{77}{36}$.

Let $X$ be a random variable having binomial distribution $B(7, p)$. If $P(X=3) = 5P(X=4)$,then the sum of the mean and the variance of $X$ is

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