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Probability Proof

  • Thread starter kuahji
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  • #1
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We have three events A, B, and C such that
P(A|C)[itex]\geq[/itex]P(B|C) and P(A|C')[itex]\geq[/itex]P(B|C')

Prove that P(A)[itex]\geq[/itex]P(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)[itex]\geq[/itex]P(BC)/P(C) and since P(C)[itex]\geq[/itex]0,
P(AC)[itex]\geq[/itex]P(BC).

Similarly from above,
P(AC')[itex]\geq[/itex]P(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:

Answers and Replies

  • #2
vela
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How did you get P(A|C)=P(AB)/P(C)? Why is B there?
 
  • #3
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How did you get P(A|C)=P(AB)/P(C)? Why is B there?
Typo, fixed it above.
 
  • #4
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If I'm not mistaken then P(A and C) + P(A and not C) = P(A).
 
  • #5
HallsofIvy
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We have three events A, B, and C such that
P(A|C)[itex]\geq[/itex]P(B|C) and P(A|C')[itex]\geq[/itex]P(B|C')

Prove that P(A)[itex]\geq[/itex]P(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)[itex]\geq[/itex]P(BC)/P(C) and since P(C)[itex]\geq[/itex]0,
P(AC)[itex]\geq[/itex]P(BC).

Similarly from above,
P(AC')[itex]\geq[/itex]P(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 [itex]P(A\cup C)= P(A)+ P(C)- P(A\cap C)[/itex]

with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.
 
  • #6
394
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Okay, just figured it out. Thanks everyone for the assistance.
 
Last edited:

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