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September 15th, 2019, 08:00 AM   #1
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Probability inequality problem

Suppose A and B are independent and identically distributed.
For the purposes of this problem, x and y are not capped at 1, but are definitely at least 0.

Pr(A)<x
Pr(B)<y
Pr(AUB)<?


I know Pr(AUB)<x+y, but is it possible to make the right side any lower? What is the lowest it can be given the information above?
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September 15th, 2019, 01:51 PM   #2
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$Pr(A\cup B)=Pr(A)+Pr(B)-Pr(A\cap B)=Pr(A)+Pr(B)-Pr(A)Pr(B)$. You are now on your own.
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September 16th, 2019, 07:29 AM   #3
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This is one of those, "It's so obvious once you see the solution," problems.
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