Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
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'):
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.