PDA

View Full Version : Proving Probabilities

hollysti
03-24-2008, 10:30 AM
I am supposed to show that if P(B) is not equal to 0, then P(A|B) is greater than or equal to 0, P(B|B) = 1, and P(A1 union A2 union A3...|B) = P(A1|B) + P(A2|B) + P(A3|B) + ... for any sequence of mutually exculsive events A1, A2, A3...

It doesn't seem like it should be so hard... in fact, it makes perfect sense that P(B|B) = 1... but I really don't know how to show it. Please help!

tkhunny
03-24-2008, 10:45 AM
Do you have "Bayes' Theorem"?

hollysti
03-24-2008, 11:11 AM
It looks familiar, although I am not sure that it has been presented to us in class.
Ok so using the theorem P(A|B) = [(P(B|A))(P(A)) / P(B)],
since P(B) does not equal 0, then that also implies that P(B|A) does not equal 0, right? If so, then P(A|B) would be greater than or equal to 0, but I am not sure how to show that...
I think using that theorem helps a little, but I still don't have a clear picure in my mind. Could you maybe help a little more? Thanks!

pka
03-24-2008, 11:24 AM
Actually your text should have put the following relation:
P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P(B)}} or P\left( {A \cap B} \right) = P\left( {A|B} \right)P(B).
In either form the proofs follow at once if you note P\left( {A \cap B} \right) \ge 0.

hollysti
03-24-2008, 11:40 AM
Ok, actually, I have that written as the first line of my proof, but my book defines that as P(B|A), not P(A|B). Are they the same? If so, can I just note that P(A union B) is greater than or equal to 0? Can you help me with the first one so I see what it looks like?

pka
03-24-2008, 11:48 AM
For any two events E&F
P\left( {E\cap F}\right) = P(E|F)P(F)=(F|E)P(E).
As you can see they are not the same.

hollysti
03-24-2008, 12:04 PM
Ok, sorry, you are right. I had written it down wrong. But I am still not sure how to show that P(A union B) is greater than or equal to 0.

hollysti
03-24-2008, 12:11 PM
Never mind I think I've got it. Thank you very much!