Continue to Site

Welcome to EDAboard.com

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.

quick probability question

Status
Not open for further replies.

oermens

Advanced Member level 2
Advanced Member level 2
Joined
Nov 19, 2005
Messages
525
Helped
107
Reputation
220
Reaction score
44
Trophy points
1,308
Location
canada
Activity points
3,988
in conditional probablilty

P(B/A) = 1 - P(B/notA) ?
 

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.
 

    oermens

    Points: 2
    Helpful Answer Positive Rating
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

    Points: 2
    Helpful Answer Positive Rating
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?
 

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.
 

    oermens

    Points: 2
    Helpful Answer Positive Rating
I think this solution P(A'|B)=1-P(A|B) is correct...if you draw a venn diagram it is better realizable.
 

    oermens

    Points: 2
    Helpful Answer Positive Rating
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'):

P(B/A')=(P(B)-P(B/A)P(A))/(1-P(A)).
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top