Proving P(A)≥P(B): Probability Proof

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Homework Help Overview

The discussion revolves around proving a probability inequality involving three events A, B, and C, specifically that P(A) is greater than or equal to P(B) given certain conditional probabilities. Participants are exploring the relationships between these probabilities and the implications of the conditions provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivation of conditional probabilities and their implications. There are attempts to manipulate these probabilities to reach the desired inequality. Questions arise regarding the correctness of expressions used and the relationships between events.

Discussion Status

The discussion includes various attempts to clarify the expressions for conditional probabilities and their expansions. Some participants have identified potential errors in the formulations, while others are actively seeking guidance on how to proceed with the proof. There is a mix of exploration and clarification without a clear consensus on the next steps.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available and the methods they can use. There are also indications of confusion regarding the correct application of probability rules.

kuahji
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We have three events A, B, and C such that
P(A|C)\geqP(B|C) and P(A|C')\geqP(B|C')

Prove that P(A)\geqP(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)\geqP(BC)/P(C) and since P(C)\geq0,
P(AC)\geqP(BC).

Similarly from above,
P(AC')\geqP(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC) with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.
 
Last edited:
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How did you get P(A|C)=P(AB)/P(C)? Why is B there?
 
vela said:
How did you get P(A|C)=P(AB)/P(C)? Why is B there?

Typo, fixed it above.
 
If I'm not mistaken then P(A and C) + P(A and not C) = P(A).
 
kuahji said:
We have three events A, B, and C such that
P(A|C)\geqP(B|C) and P(A|C')\geqP(B|C')

Prove that P(A)\geqP(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)\geqP(BC)/P(C) and since P(C)\geq0,
P(AC)\geqP(BC).

Similarly from above,
P(AC')\geqP(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC)
You have this backwards. It should be P(A\cup C)= P(A)+ P(C)- P(A\cap C)

with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.
 
Okay, just figured it out. Thanks everyone for the assistance.
 
Last edited:

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