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Homework Help: Probability question on jobs

  1. Jan 20, 2005 #1


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    Two firms V and W consider bidding on a road-building job, which may or may not be awarded depending on the amounts of the bids. Firm V submits a bid and the probability is 3/4 that it will get the job provided firm W does not bid. The probability is 3/4 that W will bid and if it does, the probability that V will get the job is only 1/3.
    1. What is the probability that V will get the job?
    2. If V gets the job, what is the probability that W did not bid?

    I spoke to my professor on how I should do this question and all she would tell me was that when we're given a lot of information in the question, apply Bayes' Theorem but I'm really not sure how to do that. I've never taken any courses in probability or stats before so any suggestions or ideas on how I could start off the question would be greatly appreciated. Thanks =)
  2. jcsd
  3. Jan 20, 2005 #2
    Problems like these aren't completely dissimilar to the basic counting approach to probability, i.e. # of ways to succeed / # of ways total.

    Company W has a 3/4 probability of bidding. We have two cases.

    Case 1: Company W bids. Chance of this happening: 3/4

    Case 2: Company W does not bid. Chance of this happening : 1/4

    Within Case 1, there are again two cases.

    Case 1: Company W bids (3/4):
    ----Case a: Company V wins the bid (1/3)
    ----Case b: Company V does not win the bid (2/3)

    Within case 2, there are also two cases.

    Case 2: Company W does not bid (1/4):
    ----Case a: Company V wins the bid (3/4)
    ----Case b: Company V does not win the bid (1/4)

    So in what cases does Company V get the bid? Cases 1.a and 2.a. So we add the probabilities of these cases happening.

    P(1.a) = P(1)*P(1.a) = (3/4)*(1/3)
    P(2.a) = P(2)*P(2.a) = (1/4)*(3/4)

    P(V wins the bid) = P(1.a) + P(2.a) = 3/4*1/3 + 1/4*3/4 = 3/4(7/12) = 7/16

    For part b, it is assumed that V wins the bid. So Cases 1.b and 2.b clearly did not happen. This changes the total probability, analagous to reducing the (total # of ways). Now the only two viable options are Cases 1.a and 2.a, so the total probability is P(1.a) + P(2.a). The probability of Company W not bidding is exhibited by Case 2.a, so the probability that Company W does not bid given that Company V wins the bid will be P(2.1)/(P(1.a) + P(2.a)).

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