Question #91983

A number x is selected at random in the interval [-1, 2]. Let the events A={x<0}, B={|x-0.5|<0.5}, and C=x>0.75. Find P[A|B], P[B|C], P[A|C^c] and P[B|C^c].

Expert's answer

Assume that the probability of any subinterval I of [-1, 2] is proportional to its length


P(I)=k⋅length(I)P(I)=k \cdot length(I)

If we let I=[−2,1],I=[-2, 1], then we must have


1=P(S)=P([−2,1])=k⋅length([−2,1])=k⋅31=P(S)=P([-2, 1])=k\cdot length([-2,1])=k\cdot 3k=13k={1 \over 3}P[A]=13length([−1,0))=13⋅1=13P[A]={1 \over 3}length([-1,0))={1 \over 3}\cdot 1={1 \over 3}

∣x−0.5∣<0.5=>−0.5<x−0.5<0.5=>0<x<1|x-0.5|<0.5=>-0.5<x-0.5<0.5=>0<x<1


P[B]=13length((0,1))=13⋅1=13P[B]={1 \over 3}length((0,1))={1 \over 3}\cdot 1={1 \over 3}

P[C]=13length((0.75,2))=13⋅1.25=512P[C]={1 \over 3}length((0.75,2))={1 \over 3}\cdot 1.25={5 \over 12}

P[A∣B]=P[A∩B]P[B]=0P[A|B]={P[A\cap B] \over P[B]}=0

P[B∩C]=13length((0.75,1))=13⋅0.25=112P[B\cap C]={1 \over 3}length((0.75,1))={1 \over 3}\cdot 0.25={1 \over 12}

P[B∣C]=P[B∩C]P[C]=112512=15P[B|C]={P[B\cap C] \over P[C]}={{1 \over 12} \over {5 \over 12}}={1 \over 5}

P[CC]=1−512=712P[C^C]=1-{5 \over 12}={7 \over 12}

P[A∩CC]=13length([−1,0))=13⋅1=13P[A\cap C^C]={1 \over 3}length([-1,0))={1 \over 3}\cdot 1={1 \over 3}


P[A∣CC]=P[A∩CC]P[CC]=13712=47P[A|C^C]={P[A\cap C^C] \over P[C^C]}={{1 \over 3} \over {7 \over 12}}={4 \over 7}

P[B∩CC]=13length(0,0.75])=13⋅0.75=14P[B\cap C^C]={1 \over 3}length(0,0.75])={1 \over 3}\cdot 0.75={1 \over 4}

P[B∣CC]=P[B∩CC]P[CC]=14712=37P[B|C^C]={P[B\cap C^C] \over P[C^C]}={{1 \over 4} \over {7 \over 12}}={3 \over 7}


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