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Question Number 105812 by Ar Brandon last updated on 31/Jul/20

$$\mathrm{In}\mathrm{a}\mathrm{by}-\mathrm{election}20\mathrm{\%}\mathrm{of}\mathrm{the}\mathrm{electorate}\mathrm{voted}\mathrm{for}\mathrm{Mr}\mathrm{X}.$$$$\mathrm{If}5\mathrm{voters}\mathrm{are}\mathrm{chosen}\mathrm{at}\mathrm{random}\mathrm{from}\mathrm{this}\mathrm{electorate},\mathrm{what}$$$$\mathrm{is}\mathrm{the}\mathrm{probability}\mathrm{that}20\mathrm{\%}\mathrm{of}\mathrm{the}\mathrm{sample}\mathrm{voted}\mathrm{for}\mathrm{Mr}\mathrm{X}?$$$$\left(\mathrm{a}\right)1\left(\mathrm{b}\right)\frac{1}{5}\left(\mathrm{c}\right)\frac{1}{625}\left(\mathrm{d}\right)\frac{256}{625}$$

Answered by Don08q last updated on 31/Jul/20

$$a=\mathrm{P}\left(\mathrm{X}\right)=\frac{20}{100}$$$$=\frac{1}{5}$$$$b=\mathrm{P}\left(\stackrel{-}{\mathrm{X}}\right)=1-\frac{20}{100}$$$$=\frac{4}{5}$$$$20\mathrm{\%}of5voters=1$$$$\mathrm{P}(\mathrm{X}=1)={}^5{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^4$$$$=\frac{256}{625}$$$$Theansweris\mathrm{D}$$

Commented by Ar Brandon last updated on 31/Jul/20

$$\mathrm{Thanks}$$

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