Question with intersects and complements.

  • Context: Undergrad 
  • Thread starter Thread starter Shawj02
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Discussion Overview

The discussion revolves around a probability problem involving intersections and complements of events A, B, and C. Participants explore how to calculate the probability of the union of intersections given specific probabilities and conditions.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states that since P(AnBnC)=0, the events are mutually exclusive, leading to the conclusion that P(AnB)=0, P(AnC)=0, and P(BnC)=0, which seems incorrect as it would imply P((AnB)U(AnC))=0.
  • Another participant proposes a formula: P(A) - P(AnB^c) + P(A) - P(AnC^c) = P((AnB)U(AnC) and seeks confirmation on its correctness.
  • A later reply confirms the formula is correct and asks if the original poster understands how it works, suggesting that a numerical result can be derived from it.

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints on how to approach the problem, with some participants agreeing on the formula while others express uncertainty about the implications of their calculations. No consensus is reached on the overall interpretation of the probabilities involved.

Contextual Notes

Participants express concerns about how to manipulate given probabilities, indicating potential limitations in understanding the relationships between the events and their complements.

Shawj02
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Ok, first post.
So I have this question, which goes something like this...
Given
P(A)=0.3, P(B)=0.3, P(C)=0.7
P(AnB^c)=0.2, P(AnC^c)=0.2, P(AnBnC)=0

Find P((AnB)U(AnC))

(Where; n =intersect, U union, ^c = complement.)

Personally my thoughts are..
P(AnBnC)=0. Therefore mutually exclusive.

And then Because probability cannot be negative. I think that leads to P(AnB)=0,P(AnC)=0 & p(BnC)=0.
Which couldn't be right, As that would give P((AnB)U(AnC)) = 0U0 = 0.

My major concern is how do I change P(AnC^c) & P(AnB^c) to something useful!

Thanks!
 
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Ive got
P(A)- P(AnB^c) + P(A)- P(AnC^c)= P((AnB)U(AnC))

Anyone want to double check me?
 
Shawj02 said:
Ive got
P(A)- P(AnB^c) + P(A)- P(AnC^c)= P((AnB)U(AnC))

Anyone want to double check me?


It's correct. Do you understand how? Also you obviously can get a number from it.
 
Awesome. Yeah, I understand how. Just had a block a guess.
Thanks.
 

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