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Let \mathrm{X} have a binomial distribution \mathrm{B(n, p)} such that the sum and the product of the mean and variance of \mathrm{X} are 24 and 128 respectively. If \mathrm{P}(\mathrm{X}>\mathrm{n}-3)=\frac{\mathrm{k}}{2^{\mathrm{n}}}, then \mathrm{k} is equal to :

Option: 1

528


Option: 2

529


Option: 3

629


Option: 4

630


Answers (1)

best_answer

\begin{aligned} & \mathrm{n p+n p q=24 \Rightarrow m p(1+q)=24 }\\ &\Rightarrow \mathrm{ n p(2-p)=24 } \\ &\Rightarrow \mathrm{n p=\frac{24}{2-p}}\\ & \text{Also} \mathrm{ n p \cdot n p q=128}\\ & \Rightarrow \mathrm{n^{2} p^{2} q=128 }\\ & \Rightarrow \mathrm{n^{2} p^{2}(1-p)=128} \\ & \Rightarrow \mathrm{ \frac{24^{2}}{(2-p)^{2}}(1-p)=128 }\\ & \Rightarrow \mathrm{\frac{(1-p)}{(2-p)^{2}}=\frac{2}{9}} \\ & \Rightarrow \mathrm{9-9 p=2\left(p^{2}+4-4 p\right)} \\ & \Rightarrow \mathrm{2 p^{2}-8 p+8+9 p-9=0 }\\ & \Rightarrow \mathrm{2 p^{2}+p-1=0} \\ & \Rightarrow \mathrm{p=\frac{1}{2} \quad(\operatorname{as} p>0)} \\ & \Rightarrow \mathrm{n=32} \\ & \therefore \mathrm{P(X>n-3) }\\ & \mathrm{=P(X>29) }\\ & \mathrm{=P(30)+P(31)+P(32)} \\ \end{aligned}\begin{aligned} & \mathrm{=^{32}C_{30} \cdot \frac{1}{2^{32}} + ^{32}C_{31} \cdot \frac{1}{2^{32}}+^{32}C_{32} \cdot \frac{1}{2^{32}}}\\ &=\mathrm{\frac{529}{2^{32}} }\\ &\therefore \quad \mathrm{k=529} \\ &\therefore \text { option (B) } \\ \end{aligned}

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Divya Prakash Singh

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