Three contractors are bidding for a project, and the client has randomly selected one to receive...

Three contractors are bidding for a project, and the client has randomly selected one to receive the contract. Contractor 2 secretly asks the client to divulge only to him one of his competitors that will not receive the contract. The client is concerned that divulging this information would increase Contractor 2's probability of receiving the contract from one-third to hone-half. Is the client's logic sound?

I know Bayes Theorem is required for this question but I'm not sure how to apply it.

Expert Solution

Let three contractors be A, B and C.

Let A, B and C be the events that contractor 1, 2 or 3 receive the contract respectively.

Then P(A) = P(B) = P(C) = 1/3

Let NC be the event that the client told to contractor 2 that contractor 1 will not receive the bid.

Given that probability that the contractor 1 will receive the bid, probability that client told to contractor 2 that contractor 1 will not receive the bid = P(NC | A) = 0

Given that probability that the contractor 2 will receive the bid, probability that client told to contractor 2 that contractor 1 will not receive the bid = P(NC | B) = 1/2 (as there are two contractors A and C which will not receive the bid)

Given that probability that the contractor 3 will receive the bid, probability that client told to contractor 2 that contractor 1 will not receive the bid = P(NC | C) = 1 (definitely, client will not inform B that he will not receive the bid and will always take contractor 1 name).

By law of total probability,

P(NC) = P(NC | A) P(A) + P(NC | B) P(B) + P(NC | C) P(C)

= 0 * (1/3) + (1/2) * (1/3) + 1 * (1/3) = 1/2

By Bayes theorem,

P(B | NC) = P(NC | B) P(B) / P(NC) = (1/2) * (1/3) / (1/2) = 1/3

Thus, given that the client told to contractor 2 that contractor 1 will not receive the bid, probability that the contractor 2 will received the bid is still 1/3.

orchestra answered 2 years ago

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