Solve mcq no. 23

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{P}\left(\mathrm{X}=\mathrm{r}\right)={\mathrm{k\alpha }}^{\mathrm{r}}, \mathrm{where} \mathrm{k} \mathrm{is} \mathrm{contant} \\ \Rightarrow \mathrm{P}\left(\mathrm{X}=0\right)={\mathrm{k\alpha }}^0 \\ \Rightarrow \mathrm{P}\left(\mathrm{X}=0\right)=\mathrm{k}....\left(\mathrm{i}\right) \\ \mathrm{Now} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{total} \mathrm{probabilty} \mathrm{is} 1, \mathrm{hence} \\ \mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=3\right)+.....+\mathrm{P}\left(\mathrm{X}=\infty \right)=1 \\ \Rightarrow \mathrm{k}+\mathrm{k\alpha }+{\mathrm{k\alpha }}^2+{\mathrm{k\alpha }}^3+..... \mathrm{to} \mathrm{infinity}=1 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{is} \mathrm{a} \mathrm{G}.\mathrm{P}. \mathrm{of} \mathrm{infinite} \mathrm{terms} \mathrm{with} \mathrm{common} \mathrm{ratio} \frac{\mathrm{k\alpha }}{\mathrm{k}}=\mathrm{\alpha } \\ \mathrm{As} 0<\mathrm{\alpha }<1, \mathrm{hence} \mathrm{sum} \mathrm{of} \mathrm{infinite} \mathrm{terms} \mathrm{of} \mathrm{G}.\mathrm{P}. \mathrm{can} \mathrm{be} \mathrm{applied} \\ \Rightarrow \frac{\mathrm{First} \mathrm{Term}}{1-\mathrm{Common} \mathrm{Ratio}}=1 \\ \Rightarrow \frac{\mathrm{k}}{1-\mathrm{\alpha }}=1 \\ \Rightarrow \mathrm{k}=1-\mathrm{\alpha } \\ \mathrm{Using} \left(\mathrm{i}\right) \mathrm{we} \mathrm{get} \\ \mathrm{P}\left(\mathrm{X}=0\right)=1-\mathrm{\alpha }...\mathrm{Option}\left(\mathrm{a}\right) \end{gathered}$$

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