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Question Number 7122 by 314159 last updated on 11/Aug/16

Answered by uchechukwu okorie favour last updated on 11/Aug/16

let the probability that a student  picked at random be taller than  1.8 metres be Pr(T)  Also; let the probability that a  student picked at random wear a  spectacle be Pr(S)  Then the probability that the   student is not taller than 1.8 metres  is Pr(T′)  Also the probability that the   student is not wearing spectacle  is Pr(S^′ )  Pr(T)=0.2=(2/(10))  Pr(T^′ )=1−0.2=0.8=(8/(10))  Pr(S)=0.3=(3/(10))  Pr(S^′ )=1−0.3=0.7=(7/(10))  let probability of the event   happening be Pr(E)  Pr(E)=Pr(TTT^′ )∪Pr(SS^′ S^′ )  Pr(E)=((2/(10))×(2/(10))×(8/(10)))∪Pr((3/(10))×(7/(10))×(7/(10)))  Pr(E)=((32)/(1000))+((147)/(1000))  Pr(E)=((179)/(1000))

$${let}\:{the}\:{probability}\:{that}\:{a}\:{student} \\ $$$${picked}\:{at}\:{random}\:{be}\:{taller}\:{than} \\ $$$$\mathrm{1}.\mathrm{8}\:{metres}\:{be}\:{Pr}\left({T}\right) \\ $$$${Also};\:{let}\:{the}\:{probability}\:{that}\:{a} \\ $$$${student}\:{picked}\:{at}\:{random}\:{wear}\:{a} \\ $$$${spectacle}\:{be}\:{Pr}\left({S}\right) \\ $$$${Then}\:{the}\:{probability}\:{that}\:{the}\: \\ $$$${student}\:{is}\:{not}\:{taller}\:{than}\:\mathrm{1}.\mathrm{8}\:{metres} \\ $$$${is}\:{Pr}\left({T}'\right) \\ $$$${Also}\:{the}\:{probability}\:{that}\:{the}\: \\ $$$${student}\:{is}\:{not}\:{wearing}\:{spectacle} \\ $$$${is}\:{Pr}\left({S}^{'} \right) \\ $$$${Pr}\left({T}\right)=\mathrm{0}.\mathrm{2}=\frac{\mathrm{2}}{\mathrm{10}} \\ $$$${Pr}\left({T}^{'} \right)=\mathrm{1}−\mathrm{0}.\mathrm{2}=\mathrm{0}.\mathrm{8}=\frac{\mathrm{8}}{\mathrm{10}} \\ $$$${Pr}\left({S}\right)=\mathrm{0}.\mathrm{3}=\frac{\mathrm{3}}{\mathrm{10}} \\ $$$${Pr}\left({S}^{'} \right)=\mathrm{1}−\mathrm{0}.\mathrm{3}=\mathrm{0}.\mathrm{7}=\frac{\mathrm{7}}{\mathrm{10}} \\ $$$${let}\:{probability}\:{of}\:{the}\:{event}\: \\ $$$${happening}\:{be}\:{Pr}\left({E}\right) \\ $$$${Pr}\left({E}\right)={Pr}\left({TTT}^{'} \right)\cup{Pr}\left({SS}^{'} {S}^{'} \right) \\ $$$${Pr}\left({E}\right)=\left(\frac{\mathrm{2}}{\mathrm{10}}×\frac{\mathrm{2}}{\mathrm{10}}×\frac{\mathrm{8}}{\mathrm{10}}\right)\cup{Pr}\left(\frac{\mathrm{3}}{\mathrm{10}}×\frac{\mathrm{7}}{\mathrm{10}}×\frac{\mathrm{7}}{\mathrm{10}}\right) \\ $$$${Pr}\left({E}\right)=\frac{\mathrm{32}}{\mathrm{1000}}+\frac{\mathrm{147}}{\mathrm{1000}} \\ $$$${Pr}\left({E}\right)=\frac{\mathrm{179}}{\mathrm{1000}} \\ $$

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