Question #178486

a)   A company has taken a ballot of its workforce, asking them to choose between two alternative new schemes (A and B) for claiming travelling expenses. 60% of the workforce support scheme A, while 50% of these owning a car. If only 30% of those who support scheme B own a car, calculate the probability that if a worker is chosen at random, he will be:

    i.       a car owner

  ii.       a car owning, scheme A supporter

 iii.       a car owner, given that he supports scheme A

 iv.       a scheme A supporter, given that he owns a car



1
Expert's answer
2021-05-07T09:27:59-0400

Probability of workforce who support scheme A P(A)=0.6P(A)=0.6

Probability of workers who support scheme B P(B)=0.4P(B)=0.4


Let X denote that person owns a car-


Probability that person support A scheme and owns a car P(AX)=0.5×0.6=0.30P(\dfrac{A}{X})=0.5\times 0.6=0.30


Probability that person support B scheme and owns a car P(BX)=0.3×0.4=0.12P(\dfrac{B}{X})=0.3\times 0.4=0.12


(a) Probability that a worker is car owner P(X)=P(AX)+PBX)=0.30+0.12=0.42P(X)=P(\dfrac{A}{X})+P\dfrac{B}{X})=0.30+0.12=0.42


(b) Probability that worker owns a car and support A schme P(AX)=0.30P(\dfrac{A}{X})=0.30


(c) Probability that worker is a car owner, given that he supports scheme A

=P(X)P(A)=0.420.6=0.7=\dfrac{P(X)}{P(A)}=\dfrac{0.42}{0.6}=0.7


(d) Probability that woker is schemeA , suppoter given that he owns a car=P(A)P(X)=0.60.42=0.3=\dfrac{P(A)}{P(X)}={0.6}{0.42}=0.3


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