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'):
P(B/A')=(P(B)-P(B/A)P(A))/(1-P(A)).