# Probability Exercise - Contracts

Here I have this exercise which I am not sure of how to approach, it is in the conditional probability section but I cannot see the use in here, I will state the question and then state my intuitive approach, so:

1. Consider four computer firms A, B, C, D bidding for a certain contract. A survey of past bidding success of these firms show the following probabilities of winning:

P(A) = 0.35, P(B) = 0.15, P(C) = 0.3, P(D) = 0.2

Before the decision is made to award the contract, firm B withdraws the bid. Find the new probabilities of winning the bid for A, C and D.

So intuitively, the exit of the company B will affect the P(A), P(C) and P(D) but not the overall probability which is P(S) = 1.

I am thinking of this solution but I might be wrong, it works and the sum of the events adds up to 1 so... Let me know, here it is:

P(A|B) = P(A) + $\frac{P(B)}{3}$ = 0.40

P(C|B) = P(C) + $\frac{P(B)}{3}$ = 0.35

P(D|B) = P(D) + $\frac{P(B)}{3}$ = 0.25

So P(A|B) + P(C|B) + P(D|B) = 0.4 + 0.35 + 0.25 = 1

I know it's wrong mathematically speaking, I mean in using of the theorems, but the result is correct, but I don't feel like it is the way to go. How am I supposed to use these three theorems here?:

P(B|A) = $\frac{P(A \cap B)}{P(A)}$

and

Plus can anyone explain to me this image (above) is it either the top or bottom?

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The problem isn't absolutely clear about what "firm B withdraws the bid" means. Are you interested in possible meanings for this sentence or just want to know what was probably intended by the person who wrote the question (taking into account it's in conditional probability). –  madprob Jan 8 at 19:00
By the way, the image means BOTH top and bottom. That is, $P(A \cap B) = P(A)P(B|A) = P(B)P(A|B)$. –  madprob Jan 8 at 19:02
They intend you to divide the probabilities for A, C, D by $0.85$. That is reasonable only under unreasonable assumptions. –  André Nicolas Jan 8 at 19:06
@madprob both would be perfect, I can't seem to grasp this problem... :( –  Chris Dobkowski Jan 8 at 19:14

Let $N$ be the event that B does not get the contract. Let $A$ be the event that A gets the contract. We calculate $\Pr(A|N)$. Please note that this is does not solve the problem. The conditional probability $\Pr(A|N)$ is not necessarily the probability described in the problem. But it is very likely to be what you are intended to compute.

By the usual defining formula for conditional probability, we have $$\Pr(A|N)=\frac{\Pr(A\cap N)}{\Pr(N)}.$$ Since whenever $A$ happens, it must be that $N$ happens, we have $\Pr(A\cap N)=\Pr(A)$. Thus our conditional probability is $\frac{0.35}{0.85}$. Similar calculations can be made for the other two required probabilities.

Remark: Suppose that A and B are non-union shops, and C and D are union shops. It is perfectly possible that the current project comes from a rabidly anti-union company. So if B withdraws, they will go to A.

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Do we really need to use conditional probability here? Assume the probabilities are the result of $100$ past bidding contracts. Hence the number of bidding contracts won by each firm, out of a total of $100$, is: $A=35,B=15,C=30,D=20$. Now since $B$ is out of the new bid, the survey is done again,but this time using only the result of past contracts where $B$ hadn't won, i.e $85$ bidding contracts, resulting in the new probabilities:$P(A)=\frac{35}{85}=\frac{7}{17},P(C)=\frac{30}{85}=\frac{6}{17},P(D)=\frac{20}{85}=\frac{4}{17}$

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Intuitively, it's simply ratios:

P(A) = 0.35, P(B) = 0.15, P(C) = 0.3, P(D) = 0.2

B drops so...

0.35 + 0.3 + 0.2 = 0.85 (this is the "whole thing" so we just need to scale to 1.0)

0.35/0.85 + 0.3/0.85 + 0.2/0.85 = 0.85/0.85 = 1

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