Category: Science > Math
Asked by: mms03-ga
List Price: $4.00
08 Dec 2002 18:36 PST
Expires: 07 Jan 2003 18:36 PST
Question ID: 121595
If M and N are independent events, P(M,N) = P(M)P(N) then prove that M' and N' are independent.
Answered By: dannidin-ga on 09 Dec 2002 01:05 PST
This is a repost of the answer I wrote as a request for clarification: M and N are independent, therefore P(M,N) = P(M)P(N) - this is simply the definition of independent events. Now let us prove that P(M',N')=P(M')P(N'), and therefore M' and N' are independent: I denote the union of two events A and B by AuB, and their intersection by (A,B). P(M',N') = P((MuN)') = 1 - P(MuN) = 1 - P(M u (N,M')) = = 1 - (P(M) + P(N,M')) = 1 - P(M) - P(N,M') = = 1 - P(M) - (P(N) - P(N,M)) = = 1 - P(M) - P(N) + P(N,M) = = 1 - P(M) - P(N) + P(M)P(N) = = (1 - P(M))(1 - P(N)) = = P(M')P(N')
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Awesome answer. It was very easy to understand the answer to the problem, with all the steps shown accurately.
From: hailstorm-ga on 08 Dec 2002 20:37 PST
Isn't this impossible for the case of M' = N'?
From: mms03-ga on 08 Dec 2002 22:03 PST
I dont see how M' will ever equal N' If M and N are independent events, we should be able to prove that M' and N' are also independent.
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