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Pra
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Posted on 09-24-07 8:14
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A and B are discrete events. You are given values for P(A) and P(B). What value of P(NOT A NOT B) will make A and B independent? Could somebody help me to solve this problem? Thank you.
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perrybhadra
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Posted on 09-24-07 8:38
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P(not A not B)=1-P(AB)-P(notA ,B)-P(A,not B) If A and B are to be independent P(not A not B)=1-P(A)P(B)-P(not A)P(B)-P(A)P(not B) I think this should be the answer
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Guest4
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Posted on 09-24-07 8:52
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Perrybhadra is right, but if it helps you understand better, here is another way: P(not A and not B) =1- P (A union B) = 1 - [ P(A) + P(B) - P(A and B) = 1- P(A) -A(B) + P(A)P(B) (if A and B are independent) *you should get the same answer both ways.
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Pra
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Posted on 09-24-07 9:25
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Pra
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Posted on 09-24-07 9:32
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Do you guys know how to solve this one? A,B,C and D are discrete events. P(AB+CD) = P(AB) + P(CD); P(BD) = 0.6; P(NOTA BD) = 0.4. Using the Veitch diagram, find P(AB NOTC D). Thank you.
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Pra
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Posted on 09-24-07 1:48
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I solved the second one already. I got 0.2 for answer. P(B|NOTA) = ??? Given P(A), P(B) and P(AB). A and B are not independent.
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Guest4
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Posted on 09-24-07 2:12
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Seems like you need to read the chapter few more times, Pra. :) These questions ask for basic definitions in probability. If you are not clear with it, may be, it would be helpful for you to go over few more times. P(B/notA) = P(B and notA)/P(notA) =[P(B) - P(A and B)]/[1-P(A)]
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Pra
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Posted on 09-24-07 3:05
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I took probability class 4 yrs ago and now it is the base for constraint theory class. Professor gave us these problems without explaining anything and it’s not covered on the book either. I don’t have my old books anymore. Thanks for your help.
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