SOLUTION: A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning co
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Question 59039: A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A given that it wins contract B.
a) What is the probability that the company will win both contracts?
b) What is the probability that the company will win at least one of the two contracts?
c) If the company wins contract B, what is the probability that it will not win contract A?
d) What is the probability that the company will win at most one of the two contracts?
e) What is the probability that the company will win neither contract?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A given that it wins contract B.
P(A)=0.6; P(not A)=0.4
P(B)=0.5; P(not B)=0.5
P(A|B)=0.8; P(not A|B)=0.2
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a) What is the probability that the company will win both contracts?
P(A and B)= P(A|B)*P(B)=0.8*0.5=0.4
Comment:
Notice that P(A and B) is not equal to P(A)*P(B)
because 0.4 is not equal to 0.6*0.5=0.3
Therefore events A and B are dependent.
b) What is the probability that the company will win at least one of the two contracts?
P(at least one)= 1 - P(win none)= 1- 0.4*0.5=0.8
c) If the company wins contract B, what is the probability that it will not win contract A?
P(notA |B)= P(not A and B)/P(B) = 0.4*0.5/0.5 = 0.4
d) What is the probability that the company will win at most one of the two contracts?
P(at most one) = 1 - P(both)= 1-0.4 = 0.6
e) What is the probability that the company will win neither contract?
P(not A and notB)=0.4*0.5 = 0.2
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Cheers,
Stan H.
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