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

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Discussion Overview

The discussion revolves around determining whether events are mutually exclusive and whether they form a partition of a sample space, specifically when probabilities are not provided. Participants explore the definitions and implications of mutually exclusive events and partitions in the context of probability theory.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that events A, B, and C are mutually exclusive if their intersection is empty, specifically stating P(A∩B∩C) = P(A∩C) = ∅.
  • Others argue that without knowing individual probabilities or additional information, one cannot assume mutual exclusivity or determine if A∪B∪C forms a partition of D.
  • It is suggested that knowledge of individual (marginal) probabilities and the value of P(D) is necessary to make any conclusions about the relationships between the events.
  • Some participants express uncertainty about the implications of independence versus disjointness, noting that these concepts are distinct and affect the calculation of probabilities differently.
  • A later reply clarifies that if events are mutually independent, the sum of their probabilities would differ from the sum if they were disjoint, indicating a need for more information to ascertain their relationships.
  • Participants discuss the conditions under which P(D) could be less than the sum of the probabilities, with some suggesting this occurs when events are dependent.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to determine mutual exclusivity without probabilities. Multiple competing views remain regarding the definitions and implications of independence and disjointness in probability.

Contextual Notes

Participants note that assumptions about mutual exclusivity or independence cannot be made without additional information, highlighting the limitations of the discussion based on the provided equation A∪B∪C = D.

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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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.
 
Last edited:
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?
 
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?
 
Last edited:
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).
 
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.
 
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?
 
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.
 
Last edited:
ar! I see now! thanks!
 
  • #10
Philip Wong said:
ar! I see now! thanks!

You're welcome.
 

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