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If A is independent of B, then
P(B/A)=P(B) and P(B/notA) =P(B) and you end up with P(B)=1-P(B).
Here is an actual example. Suppose you have a random variale X taking 1,2,and 3 with the same probability 1/3. Set A={1,2}, then notA={3}, and set B={1}. Therfore,
P(B/A)=1/2, P(B/notA)=0. Therefore, P(B/A) < 1 - P(B/notA).
thanks. What is the relationship between P(A/B) and P(notA/B)? Would it be P(A) and P(notA) respectively? Here is the question I am trying I am having problem with:
A construction company has submitted bids on two separate contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A if it wins contract B.
c) If the company wins contract B, what is the probability that it will not win contract A?
Thanks for the help. I had orignally misread the question and thought it was looking for P(B/notA), so I did get P(notA/B) = 0.2 and used Bayes Thm to find P(B/notA). In any case is there a relationship between P(B/A) and P(B/notA) as in my original post or is it steve10's explanation?
Based on your description in the previous posts, I am guessing that P(notA/B) may not be what you wanted as it is trivial and does not need P(A) or P(B). In other words, you might still want to get P(B/notA). Actually, P(B/notA) is the probability that the company still wins B when it fails to win A, which makes sense.
Let's use the notation A'=notA, suggested by someone upstairs.
According to your posts, we have the following:
P(A)=0.6
P(B)=0.5
P(A/B)=0.8
Therefore, you have
P(AB)=P(A/B)P(B)=0.8*0.5=0.4
We then have
P(B/A)=P(AB)/P(A)=0.4/0.6=2/3 (=the probability that the company wins B when it wins A)
P(B/A')=P(A'B)/P(A')=(P(B)-P(AB))/(1-P(A))=(0.5-0.4)/(1-0.6)=0.25
If you want a relation between P(B/A) and P(B/A'), here you can get it.
P(AB)=P(B/A)P(A)
P(A'B)=P(B/A')P(A')
Since P(AB)+P(A'B)=P(B) and P(A')=1-P(A), we have
P(B)=P(B/A)P(A)+P(B/A')(1-P(A))
First make sure that P(A)<1 (as in our case), then you solve it for P(B/A'):
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