# Thread: Proof of Condditional Probabilities

1. ## Proof of Condditional Probabilities

I'm horrible at proofs. Can someone help me with this problem? Many thanks

Show that the postulates of probability are satisfied by conditional probabilities. IN other words, show that if P(B)≠0

1. P(A|B)≥0
2. P(B|B)=0
3. P(A1UA2U...|B)=P(A1|B)+P(A2|B)+...for any sequence of mutually exclusive events A1, A2...

2. Originally Posted by xuyuan
show that if P(B)≠0
1. P(A|B)≥0
2. P(B|B)=0
3. P(A1UA2U...|B)=P(A1|B)+P(A2|B)+...for any sequence of mutually exclusive events A1, A2...
#2 Written incorrectly. It should be $\displaystyle P(B|B)=\frac{P(B\cap B)}{P(B)}=\frac{P(B)}{P(B)}=\color{blue}1$.

#1 is self evident. Probabilites are always non-negative.

For any events if $\displaystyle A\cap B =\emptyset$ then it ia always true that $\displaystyle P(A\cup B) =P(A)+P(B)$
So $\displaystyle P[(A_1\cap B)\cup (A_2\cap B)\cup,\cdots,(A_n\cap B)]=P(A_1\cap B)+P(A_2\cap B)+\cdots P(A_n\cap B)$

3. thank you very much!

4. 1 is obvious as plato said

what I like to show is that

$\displaystyle P(A|B)\le 1$

That follows from $\displaystyle P(AB)\le P(B)$ by set containment