# Finding P(A|B)

## Main Question or Discussion Point

HI guys,
I was wondering how do find P(A|B)
To calculate the following.

P(A|B) = P(B|A)P(A) / (P(B|A)*P(A) + P(B|A')*P(A'))

I know that
P(B'|A) = .001
P(B|A') = 0.09 and
P(A') = .99

P(A) = .01
and P(B'|A') = .91

I'm pretty sure that it requires the use of P(B'|A) = .001 but I can't seem to understand the relationship.

Could some one please point me in the right direction.
regards
Brendan

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Sorry P(B|A)
Brendan

HallsofIvy
Homework Helper
Given A, either B or B' so P(B|A)= 1- P(B'|A).

I was using bayes theorem to find the probabilty of being infected given that a test was positive.
So I 've used you information about finding P(B|A) to find P(A|B).
Here's what I found.

We need to find.
P(A|B) = P(B|A)P(A) / (P(B|A)*P(A) + P(B|A')*P(A'))

We're Given A, either B or B' so P(B|A)= 1- P(B'|A) = 1 -.001

P(B|A) = .999
P(A) = .01
P(A') = .99
P(B|A')=.09

Therefore:

P(A|B) = P(.999)P(.01) / (P(.999)*P(.01) + P(.09)*P(.99))

P(A|B) = .00999/00.999 + .0891

P(A|B) = .1008

So the patient only has approx 10% chance of having the desease even after testing positive.

once again thanks for your help.

yes, that's why you need more than one test in order to provide some independence to the results; the false positives get winnowed down

HallsofIvy
Homework Helper
I was using bayes theorem to find the probabilty of being infected given that a test was positive.
So I 've used you information about finding P(B|A) to find P(A|B).
Here's what I found.

We need to find.
P(A|B) = P(B|A)P(A) / (P(B|A)*P(A) + P(B|A')*P(A'))

We're Given A, either B or B' so P(B|A)= 1- P(B'|A) = 1 -.001

P(B|A) = .999
P(A) = .01
P(A') = .99
P(B|A')=.09

Therefore:

P(A|B) = P(.999)P(.01) / (P(.999)*P(.01) + P(.09)*P(.99))

P(A|B) = .00999/00.999 + .0891

P(A|B) = .1008

So the patient only has approx 10% chance of having the desease even after testing positive.

once again thanks for your help.
I assume then that A= "has disease" and B= "tested positive"? In that case we are given "Probability that a person tests positive given that he has the disease" is 0.999, "Probability that a person has the disease (before testing)" is 0.01. I wouldn't use a "formula" for this- here's what I would do. Imagine there are 100000 people. .01 of them, 1000 people, have the disease, 99000 do not. Of the 1000 people who do have the disease .999 of them, 999, test positive, and of the 99000 people who do not have the disease, .09 of them, 8910, also test positive. So out of 999+ 8910= 9909 people who test positive, 1000 of them or 1000/9909= .1009 have the disease. My answer differs slightly from yours because of round off error.