# Thread: New to stats. Need help.

1. Ok that makes sense, I'm thinking then..

(A1∩B)U(A2∩B)U...U(Aj∩B)=1. I feel like their is a more formal way of proving this however.

2. it will be
(A1∩B)U(A2∩B)U...U(Aj∩B)=B

3. Alright so then to relate it back the original equation

∑j-m P(Aj|B)=1.

We have ∑j-m (A∩B)=B? So then how do we relate the two.

Vinux, I have another question for you if you dont mind.

Suppose {Aj}j-m is a collection of mutally exclusive, collectively exhaustive events in the sample space, and let B an event with P(B) >0. Show that ∑j-m P(Aj|B)=1.

Alright so I'm thinking P(Aj|B)= P(Aj∩ B)/P(B)

And since they are mutually exclusive A∩ B= 0?

Also, since they are collectively exhaustive AUB=S=1?
From the question,
Ai∩Aj = ∅ for i ≠ j
∪Aj = Ω (union over all Aj)
Therefore, Ai∩B and Aj∩B are mutually exclusive also, for i ≠ j
For any mutually exclusive events Ai∩B and Aj∩B, for i ≠ j,
Pr{Ai∩B} + Pr{Aj∩B} = Pr{(Ai∩B)∪(Aj∩B)}
By De Morgan's Law,
(Ai∩B)∪(Aj∩B) = (Ai∪Aj)∩B

Therefore,
∑Pr{Aj|B} = ∑Pr{Aj∩B}/Pr{B} = Pr{(∪Aj)∩B}/Pr{B} = Pr{Ω∩B}/Pr{B}
= Pr{B}/Pr{B} = 1

Alright so then to relate it back the original equation

∑j-m P(Aj|B)=1.

We have ∑j-m (A∩B)=B? So then how do we relate the two.
P(Aj|B) and P(Aj∩B) are not same..

∑j-m (A∩B)=B => ∑j-m P(Aj ∩ B)=P(B) .

6. Thanks BGM and Vinux. BGM , is {(∪Aj)∩B the same as ∑j-m (A∩B)? By that i mean is{(∪Aj)∩B .. the union of all intersecions of A and B?

And if you dont mind could you explain Pr{Ω∩B} = P(B)

7. The addition follows from the third probability axioms, assumption of σ-additivity

For any events B, it must be a subset of the sample space Ω
And B⊂Ω ⇒ Ω∩B = B

8. Sorry BGM, I had never seen the notation Ω, So Ω represents the sample space. and then Pr{(∪Aj)∩B=Pr(Ω∩B) because (UAj)∩B represents all the events in the sample space?

9. UAj = Ω because the collection of the events Aj is exhaustive
{Aj} formed a partition of Ω as they are disjoint too.

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