# How to work out this probability proof?

If Π(A) denotes the odds on A and Π(A|B) the odds on A given B, deduce that

Π(A|B) / Π(A) = P(B|A) / P(B |Ā )

I really am struggling with this this question, even though it seems pretty straightforward. Some help will be hugely appreciated.

Relevance

Hey! I thought this WAS a mathematics forum!

If you'll excuse me, I'm going to write Ā as ~A, because I don't have an equivalent character for ~B and I'm going to need one.

Now, let's simplify the whole business by considering all the combined cases possible with respect to the occurrence of events A and B: let

P(A & B) = q/(q+r+s+t)

P(A & ~B) = r/(q+r+s+t)

P(~A & B) = s/(q+r+s+t)

P(~A & ~B) = t/(q+r+s+t)

We've just reduced the model to a typical Venn diagram consisting of two partially overlapping circles inside a larger circle, with the four spaces thus created marked q, r, s, and t. We can now express odds and probabilities using these four variables.

Π(A|B) = q/s

Π(A) = (q+r)/(s+t)

P(B|A) = q/(q+r)

P(B|~A) = s/(s+t)

The proof is now just a matter of algebraic manipulation:

Π(A|B) / Π(A)

= (q/s) / [(q+r)/(s+t)]

= (q/s) * [(s+t)/(q+r)]

= [q(s+t)]/[s(q+r)]

= [q/q+r)]/[s/(s+t)]

= P(B|A) / P(B|~A)

It is pretty straightforward, but only after finding a way to express odds and probabilities, including conditional ones, in terms of a common set of variables.

• For future reference, I suggest you seek out a mathematics forums. Some good ones I recommend are :

http://www.mathhelpforum.com/math-help/

Both should have sections for Probability/Statistics.

I hope that was useful!

Source(s): I'm 1337 with my interwebs