# Basic Probability Proof

• August 31st 2012, 04:41 PM
Hal2001
Basic Probability Proof
P(A|C)≥P(B|C) and P(A|C')≥P(B|C')

Prove that P(A)≥P(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)≥P(BC)/P(C) and since P(C)≥0,

P(AC)≥P(BC).

Similarly from above,
P(AC')≥P(BC').

Also, P(AC)=P(A)+P(C)-P(AUC) which led to what is below.

As of now I've gotten down to... but got stuck (Doh)

P(A)≥P(B)-P(BUC)+P(AUC)
P(A)≥P(B)-P(BUC')+P(AUC')
• August 31st 2012, 05:06 PM
Plato
Re: Basic Probability Proof
Quote:

Originally Posted by Hal2001
P(A|C)≥P(B|C) and P(A|C')≥P(B|C')
Prove that P(A)≥P(B).

It is so easy.
From the given we have: $P(AC)>P(BC)~\&~P(AC')>P(BC')$.
But simply add those two and realize that $P(A)=P(AC)+P(AC')$.
• August 31st 2012, 05:12 PM
Hal2001
Re: Basic Probability Proof
Got it. Hmm... I really was making it much harder than it should have been. Thanks.