My Math Forum A sets complement

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 February 17th, 2014, 08:50 PM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 A sets complement Given, $P(A)= .4$ , $P(B)=.5$ , and $P(A\cap B)=.3$ Find $P(A\cup B'" /> Just visualizing this, I am sure the answer is $.1$ Is the correct set notation, and what is given, would this be $P(A \cap B'= P(A) - P(A \cap B)" /> ? thanks!
 February 18th, 2014, 02:23 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: A sets complement The set you want is $A \cap B' = A \backslash (A \cap B)$ which has area $P(A \cap B'= P(A) - P(A \cap B)" />, yes.
February 18th, 2014, 04:21 AM   #3
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Re: A sets complement

Quote:
 Originally Posted by WWRtelescoping Given, $P(A)= .4$ , $P(B)=.5$ , and $P(A\cap B)=.3$ Find $P(A\cup B'" /> Just visualizing this, I am sure the answer is $.1$ Is the correct set notation, and what is given, would this be $P(A \cap B'= P(A) - P(A \cap B)" /> ? thanks!
??? $P(A\cap B'= P(A)- P(A\cap B)" />, yes, but you originally asked about $P(A\cup B'" />

Imagine 10 objects in the "universe". There are 4 in A, 5 in B, and 3 in both. So there are 4- 3= 1 that are in A but not in B. That would be $P(A\cap B'= 1" />. There are 5- 3= 2 that are in B but not in A so 10- 2= 8 that are in A or not in B. That is $P(A\cup B'= 8" />

 February 18th, 2014, 11:15 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 A = A ? (B ? B') = (A ? B) ? (A ? B'), which is a union of disjoint sets. Hence P(A) = P(A ? B) + P(A ? B'), and so P(A ? B') = P(A) - P(A ? B) = .4 - .3 = .1. Similarly, P(A' ? B) = P(B) - P(A ? B) = .5 - .3 = .2. As A ? B' = (A' ? B)', P(A ? B') = 1 - P(A' ? B) = 1 - .2 = .8. Neither calculation needs all the information given in the question.

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