If $\mathrm{X}$ is random variable with probability distribution $P(X=k)=\frac{(k+1) c}{2^k}, k=0,1,2, \ldots$, then $P(X \geq 3)=$

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Aicharya · Accepted Answer

$ 	ext { } \begin{gathered} \because P(X=k)=\frac{(k+1) C}{2^k}, k=0,1,2, \ldots . \ \sum P(X=\mathrm{k})=\mathrm{C}\left(1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\ldotsight) \ \Rightarrow 1=C\left(1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\ldotsight) \ \Rightarrow \frac{1}{2}=C\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\ldotsight) \ \quad-\quad-\quad-\quad-\quad- \ \frac{1}{2}=C\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3} \ldotsight) \Rightarrow C=\frac{1}{4} \end{gathered} $ Now, $P(X \geq 3)=1-[P(X=0)+P(X=1)+P(X=2)]$ $ =1-\left[C+\frac{2 C}{2}+\frac{3 C}{4}ight]=1-\frac{11 C}{4}=1-\frac{11}{16}=\frac{5}{16} $

If $\mathrm{X}$ is random variable with probability distribution $P(X=k)=\frac{(k+1) c}{2^k}, k=0,1,2,…

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