# Conditional Probabilty question

• Oct 9th 2008, 08:09 AM
Creol
Conditional Probabilty question
okay the problem statement is as follows
A gene can be either type A or type B (ie note type A). A gene can also be either dominant or regressive.
If a gene is type B, then there is a 0.31 probability that it is dominant. When a gene is dominant, then there is also a 0.22 probability that it is type B.

Let A be the event that the gene is of 'type A' and let D be the even that the gene is 'dominant'. What is the probability that the gene is type A?

so what is given is P(dominant|B)=0.31 and P(B|dominant)=0.22 and that A and B are mutually exclusive

Now what I'm not entirely sure is correct is, P(regressive|B)=0.69 and P(A|dominant)=0.78

Breaking down P(dominant|B) = P(dom intersect B)/P(B) and similarly for the others. I've tried every combination I can think of for solving for P(A) and I always end up with more unknowns than equations. What should I do?

I also have P(A) = (A int R + A int D)/(A int R + A int D + B int R + B int D)
similarly for P(B) = (B int R + B int D)/(A int R + A int D + B int R + B int D)
• Oct 9th 2008, 01:55 PM
hatsoff
I don't think you can do this without additional information. This is what I'd start with:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|A)P(A)}$

There are a couple of variables here that we don't know; namely:

$P(B); P(A); P(D|A)$

However, the question implies the following:

$P(A)+P(B)=1$

Which means:

$P(B)=1-P(A)$

Which we can plug in to the original equation:

$P(B|D)=\frac{P(D|B)[1-P(A)]}{P(D|B)[1-P(A)]+P(D|A)P(A)}$

But what about $P(D|A)$? Well, let's explore the Bayesian equations for this. The first one that comes to mind is:

$P(A|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)P(B)}$

This might help, because, remember:

$P(B)=1-P(A)$

Which means that:

$P(A|D)=1-P(B|D)$

So let's plug those in:

$1-P(B|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)[1-P(A)]}$

And now let's rearrange it:

$P(D|A)=\frac{P(D|B)P(A)+P(B|D)P(D|B)-P(B|D)P(D|B)P(A)-P(D|B)}{P(A)-P(B|D)P(A)-P(A)}$

Now you should be able to plug that in and solve for $P(A)$