BestOfBid.com
Probability?
9) A company is competing for three government contracts (contract A, contract B and contract C). It has been determined that the probability of winning contract A is .35, winning contract B is .6 and contract C is .8. Assuming that winning contract A, contract B, and contract C are independent, determine. a)The probability of winning all 3 contracts. b) The probability of winning at least 1 of the 3 contracts. c)The probability of winning exactly two of the three contracts.
Public Comments
- when it comes to contracts with the gov't he trhat saves the most wins the most
- a) You want to win A and B and C, so multiply .35(.6)(.8) b) The probability of winning at least one is 1- probability of winning none. To find the probability of losing all the contracts, multiply .65(.4)(.2) because these are the probabilities of losing each contract. Then subtract your answer from 1. c) There are a few ways this can happen. Win A, win B, lose C multiply .65(6)(.2) Lose A, win B, win C multiply .35(.6)(.8) Win A, lose B, win C multiply .65(.4)(.8) Add your three answers together.
- a) (prob A) * (prob B) * (prob C) = 0.35 * 0.6 * 0.8 = 0.168 b) while the chance of NOT wining any contract is (1 - prob A)*(1 - prob B)*(1 - prob C) = 0.052 the chance to win at least 1 is (1 - 0.052) = 0.948 c) (prob A * prob B * (1-prob C))+(Prob A * prob C *(1 - prob B)) + (prob B * prob C * (1 - prob A)) = 0.466
- a) The prob of winning all three contracts is 0.35*0.6*0.8 b) the prob of winning at least 1 of the 3 contracts is the same as 1 - prob of getting no contract = 1 - (1-0.35)(1-0.9)(1-0.8) c) the prob of winning exactly two of the three contracts 0.35*0.6(1-0.8) + 0.35(1-0.6)(0.8) + (1-0.35)0.6(0.8) sorry no calculator
