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Question Number 99230 by Ar Brandon last updated on 19/Jun/20

$$\ast \mathrm{\setminus }\mathcal{I}\mathrm{n}\mathrm{a}\mathrm{certain}\mathrm{country},4\mathrm{people}\mathrm{are}\mathrm{tested}\mathrm{positive}\mathrm{for}\mathrm{the}\mathrm{covid}-19\mathrm{per}\mathrm{hour}.$$$$\mathcal{D}\mathrm{etermine}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{in}\mathrm{an}\mathrm{hour};$$$$\mathrm{a}\mathrm{\setminus }\mathrm{Exactly}2\mathrm{people}\mathrm{are}\mathrm{contaminated}$$$$\mathrm{b}\mathrm{\setminus }\mathrm{At}\mathrm{most}\mathrm{one}\mathrm{person}\mathrm{is}\mathrm{contaminated}.$$$$\mathrm{c}\mathrm{\setminus }\mathrm{Between}2\mathrm{and}5\mathrm{people}\mathrm{are}\mathrm{contaminated}.$$

Commented by john santu last updated on 20/Jun/20

$$\left(\mathrm{a}\right)\mathrm{C}(4,2)\times {\left(\frac{1}{4}\right)}^2\times {\left(\frac{3}{4}\right)}^2=6\times \frac{9}{16\times 16}=\frac{27}{108}$$

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