quick probability question

[oermens]

in conditional probablilty

P(B/A) = 1 - P(B/notA) ?

 

[neoaspilet11]

Are A and B dependent or independent events?

 

[claudiocamera]

P(B/A) = P(AB)/P(A)
P(B/notA)= [P(B) -P(AB)] / [1-P(A)]

In order to P(B/A) = 1 - P(B/notA)

it would be necessary that

P(AB)/P(A) = 1- {[P(B) -P(AB)] / [1-P(A)]}

I couldnt find a way to make this equality true, so in my opinion the answer is no.

 

[steve10]

This is obviouly incorrect.

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).

 

[oermens]

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?

 

[main_road]

Since P(A/B) + P(notA/B)=1, then the probability that it will not win A under the condition that it will win B is
P(notA/B)=1-P(A/B)=1-0.8=0.2.

 

[pmonon]

I think this solution P(A'|B)=1-P(A|B) is correct...if you draw a venn diagram it is better realizable.

 

[oermens]

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?

 

[steve10]

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)).

 

[pmonon]

Yeah