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September 15th, 2019, 08:00 AM  #1 
Newbie Joined: Sep 2019 From: New York Posts: 3 Thanks: 0  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? 
September 15th, 2019, 01:51 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,852 Thanks: 743 
$Pr(A\cup B)=Pr(A)+Pr(B)Pr(A\cap B)=Pr(A)+Pr(B)Pr(A)Pr(B)$. You are now on your own.

September 16th, 2019, 07:29 AM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 380 Thanks: 205 
This is one of those, "It's so obvious once you see the solution," problems. 

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inequality, probability, problem 
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