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  • If the probability that the random variable X takes values x is given by P(X=x) = k (x+1)3-x , x = 0, 1,2, 3, ...where k is a constant, then P(X <img alt="\geq" src="https://entrancecorne

If the probability that the random variable X takes values x is given by P(X=x) = k (x+1)3-x

, x = 0, 1,2, 3, ...where k is a constant, then P(X \geq2) is equal to

Option: 1

\frac{7}{27}


Option: 2

\frac{11}{18}


Option: 3

\frac{7}{18}


Option: 4

\frac{20}{27}


Answers (1)

best_answer

\begin{aligned} & \sum_{x=0}^{\infty} P(X=x)=1 \\ & \mathrm{k}\left(1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+4 \cdot 3^{-3}+\ldots \infty\right. \\ & \text { Let } \quad s=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \infty \\ & \frac{s}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+\ldots \infty \\ & \frac{2 \mathrm{~s}}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^2}+\ldots \\ & \frac{2 s}{3}-\frac{1}{1-\frac{1}{3}}=\frac{3}{2} \\ & \mathrm{~s}=\frac{9}{4} \\ & \text { so } \quad \mathrm{k}=\frac{4}{9} \\ & \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{x}=0)-\mathrm{P}(\mathrm{x}= \\ & =1-\frac{4}{9}\left(1+\frac{2}{3}\right) \\ & =\frac{7}{27}(\text { Option 1) }) \\ & \end{aligned}

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