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  • Let be a random variable having binomial distribution <img alt="B(7, p)$. If $P(X=3)=5 P(X=4)" src="h

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

Option: 1

\frac{105}{16}


Option: 2

\frac{7}{16}


Option: 3

\frac{77}{36}


Option: 4

\frac{49}{16}


Answers (1)

best_answer

\mathrm{P(X=3)={ }^{7} C_{3} \times p^{3} \times(1-p)^{4}} \\

\mathrm{P(X=4)={ }^{7} C_{4} \times p^{4} \times(1-p)^{3} }

\mathrm{\text { Given } P(x=3)=5 p(x=4)} \\

\mathrm{\Rightarrow 7 c_{3} \times p^{3}(1-p)^{4}=5 \times 7 c_{4} \times p^{4} \times(1-p)^{3}}

\mathrm{\Rightarrow 1-p=5 p \Rightarrow p=\frac{1}{6} }

\mathrm{Mean =n p=7 \times \frac{1}{6}=\frac{7}{6}}\\

\mathrm{Variance =n p(1-p)=7 \times \frac{1}{6} \times \frac{5}{6}=\frac{35}{36}} \\

\mathrm{Mean + Variance =\frac{7}{6}+\frac{35}{36}=\frac{77}{36}}

Hence the correct answer is option 3.

Posted by

Divya Prakash Singh

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