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February 11th, 2011, 01:07 PM  #1 
Newbie Joined: Feb 2011 Posts: 15 Thanks: 0  Statistics Homework
If I have three events, A,B and C, and I want the event that only A will occur, would the formula be this? P(A n B' n C')= P(A) P(A n B)  P(A n C)  P(A n B n C) When I do it, I keep getting a negative number, 1.03, and the answer is suppose to be 0.03. P(A) = 0.70 P(B)= 0.8 P(C) = 0.75 P(A u B) = 085 P(A u C)= 0.90 P(B u C)= 0.95 P(A u B u C)= 0.98 
February 11th, 2011, 01:33 PM  #2 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Statistics Homework
Your formula uses intersections like P(A n B) but then you list unions P(A u B), P(A u C), etc. What values did you get for the intersections? Look at the Venn diagram. You subtracted P(A n B n C) three times: once when you subtracted P(A n B), once when you subtracted P(A n C), and once again at the end. The formula should end +P(A n B n C), not P(A n B n C). This is an example of the "inclusionexclusion principle". 
February 11th, 2011, 01:43 PM  #3 
Newbie Joined: Feb 2011 Posts: 15 Thanks: 0  Re: Statistics Homework
Oh, I am sorry for the confusion. Those values I posted were the given values for the question. I am still confused though. Why would it be plus at the end? If I wanted to find the event of only A occurring, wouldn't I subtract those three intersection for the probability of A? 
February 11th, 2011, 02:07 PM  #4 
Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3  Re: Statistics Homework
The problem is that P(A n B n C) is part of P(A n B). So when you took away P(A n B) you took away P(A n B n C) as well. Unfortunately the diagram I posted doesn't make that clear.

February 13th, 2011, 01:30 PM  #5 
Newbie Joined: Feb 2011 Posts: 15 Thanks: 0  Re: Statistics Homework
Oh, okay! I thought that P(A n B) was completely separate from P(A n B n C). Thank you!


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