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? 
Thank you in advance! :)
 A: 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. 
A: I would have prefered to give a comment, but can't due to missing reputation. 
To your solution: 
It is not reasonable to divide the winning probability of B equally among the others, as their winning probability hasn't been equal before. Therefore, the one with higher probability should get a greater share of B's winning probability. 
You can see it 2 ways:


*

*With relative frequencies
With the given probabilities,you can say that if B doesn't win, A has to get a share of P(A)/(P(A)+P(C)+P(D) = 0,35 / 0,85 = 0,411764.
The same way P(C)=0,3/0,85 = 0,352941 and P(D)= 0,2/0,85 = 0,235294

*You can use conditional probabilities. The first event is "B wins" with P(B) = 0,15. Therefore "B doesn't win" = 0,85. 
The idea is to find the conditional probabilities of P(A|"B doesn't win"), P(C| "B doesn't win") and P(D|"B doesn't win"). If you draw the diagram (depth 2) you see, that e.g. P(A|"B doesn't win") * P ("B doesn't win)" = 0,35 and since P ("B doesn't win)" = 0,85 it follows P(A|"B doesn't win") = 0,35 / 0,85. Same calculation for C and D. 
A: 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:$P(A)=\frac{35}{85}=\frac{7}{17},P(C)=\frac{30}{85}=\frac{6}{17},P(D)=\frac{20}{85}=\frac{4}{17}$
A: 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
