# How to determined if events disjoin, when probaility of events is not given

1. Jan 7, 2012

### Philip Wong

Hi guys,
I'm learning about partition sets of the sample space.
I understand that events form a partition of Ω when:
1) events are mutually exclusive from each other
2) union of events adds up to Ω

My question is: how do can I determine if the events are mutually exclusive to each other, when the probability for any events are not given, AND were not explicitly determined.

for example:

A∪B∪C = D

How can I determined if the above example are mutually exclusive from each other, such that I could determine whether A∪B∪C forms a partition of D.

thanks

2. Jan 7, 2012

### SW VandeCarr

Three event sets A,B,C are disjoint if $P(A\cap B\cap C) = P(A\cap C) = \emptyset$.

and A,B,C are partitions of D if $P(A\cup B\cup C) = P(D)$

I'm not sure how you're defining omega.

Last edited: Jan 7, 2012
3. Jan 7, 2012

### Philip Wong

omega is the sample space.
secondly how can we assume $P(A\cap B\cap C) = P(A\cap C) = \emptyset$, when there is no other information support the idea? Do we form such assumption from conditional probability, such that we believe this is what exactly happened?

4. Jan 8, 2012

### SW VandeCarr

You can only know if you know the individual (marginal) probabilities and the value of P(D). You need more information if P(D) is less than the simple sum of the marginal probabilities. If the probabilities are independent, P(D) will be less than the simple sum. Do you know why?

Last edited: Jan 8, 2012
5. Jan 8, 2012

### Philip Wong

Umm, not so sure why P(D) will be less than the simple sum if probabilities are independent. But I thought P(D) will be less than the sum, if only the probabilities are dependent of each other, because we have to -P(A n B), -P(A n C), -P(B n C).

6. Jan 8, 2012

### SW VandeCarr

Do you know the difference between independent and disjoint probabilities? They are not the same. Look up the definitions of the two. You can calculate P(D) from the marginal probabilities alone if they are mutually independent (and therefore not disjoint) assuming they are in same the event space.

7. Jan 8, 2012

### Philip Wong

Yup I just looked up the difference between the two and I get a better picture now. Thus I think I got my original question wrong.
So let me rephrase it to see if I get the idea right:
In order to work out if events is partition to the event space,
one of the key aspect is to see if the events are mutually exclusive (or independent) from each other.

This is what I wanted to find out originally, how can I work out if events are mutually exclusive from each other if I were only given a equation as such A∪B∪C = D?

8. Jan 8, 2012

### SW VandeCarr

The sum of three mutually disjoint probabilities is P(A)+P(B)+P(C) = P(D)

The sum of three mutually independent probabilities is P(A)+P(B)+P(C) = P(D)-(P(A)P(B)+P(A)P(C)+P(B)P(C)-P(A)P(B)P(C))

If the sum is less than on line 2 you do not have mutual independence, but you need more information to find the specific relationships.

Last edited: Jan 8, 2012
9. Jan 8, 2012

### Philip Wong

ar! I see now! thanks!

10. Jan 8, 2012

### SW VandeCarr

You're welcome.