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

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In summary: It's important to understand the difference between independent and disjoint probabilities in order to determine if events form a partition of the sample space. This is especially important in probability and statistics.
  • #1
Philip Wong
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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
 
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  • #2
Philip Wong said:
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

Three event sets A,B,C are disjoint if [itex]P(A\cap B\cap C) = P(A\cap C) = \emptyset[/itex].

and A,B,C are partitions of D if [itex]P(A\cup B\cup C) = P(D)[/itex]

I'm not sure how you're defining omega.
 
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  • #3
SW VandeCarr said:
Three event sets A,B,C are disjoint if [itex]P(A\cap B\cap C) = P(A\cap C) = \emptyset[/itex].

and A,B,C are partitions of D if [itex]P(A\cup B\cup C) = P(D)[/itex]

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

omega is the sample space.
secondly how can we assume [itex]P(A\cap B\cap C) = P(A\cap C) = \emptyset[/itex], 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
Philip Wong said:
omega is the sample space.
secondly how can we assume [itex]P(A\cap B\cap C) = P(A\cap C) = \emptyset[/itex], 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?

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?
 
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  • #5
SW VandeCarr said:
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?

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
Philip Wong said:
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).

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
SW VandeCarr said:
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.

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
Philip Wong said:
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?

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.
 
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  • #9
ar! I see now! thanks!
 
  • #10
Philip Wong said:
ar! I see now! thanks!

You're welcome.
 

1. How can I determine if events are disjoin without knowing the probability?

To determine if events are disjoin or mutually exclusive, you can use the Addition Rule. This rule states that for two events to be disjoin, their probabilities must add up to 1. If the sum of the probabilities is greater than 1, then the events are not disjoin.

2. Is it possible for events to be disjoin if their probabilities do not add up to 1?

No, if the sum of the probabilities of two events is less than 1, then they are not disjoin. This is because the remaining probability is attributed to the intersection of the events, which is not possible if the events are disjoin.

3. Can the Addition Rule be used for more than two events?

Yes, the Addition Rule can be used for any number of events. The rule states that the sum of the probabilities of all the events must equal 1 for them to be disjoin. Therefore, you can apply the same logic for more than two events.

4. How can I visually represent if events are disjoin?

You can use a Venn diagram to visually represent if events are disjoin or not. If the circles representing the events do not overlap, then the events are disjoin. If there is any overlap between the circles, then the events are not disjoin.

5. Are there any other methods for determining if events are disjoin?

Yes, you can also use the Complement Rule to determine if events are disjoin. This rule states that the probability of the complement of an event is 1 minus the probability of the event. If the complements of two events add up to 1, then the events are disjoin.

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