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 Subject: Probability Category: Science > Math Asked by: mms03-ga List Price: \$4.00 Posted: 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.``` Request for Question Clarification by maniac-ga on 08 Dec 2002 19:44 PST ```Hello Mms03, I would like to answer your question, but it appears that something is missing. The rule (which I can provide several references to) states that if M and N are independent, then the probability of M followed by N is equal to the probability of M time the probability of N. That is basically what you stated. Are you saying that P(M', N') = P(M') * P(N') [and asking for a proof they are independent] or something else? Thanks. --Maniac``` Clarification of Question by mms03-ga on 08 Dec 2002 20:05 PST ```Ill restate the question. "Based on the fact that M and N are independent, prove that M' and N' are also independent"``` Request for Question Clarification by bobby_d-ga on 08 Dec 2002 23:26 PST ```Hmm... We know M is independent of N. We can say therefore that -M + 1 is independent of -N + 1, correct? We are simply numerically manipulating M and N. but 1 - M = M' and 1 - N = N'. Putting these two facts together, M' and N' are independent. I hope that is what you are looking for. bobby_d``` Request for Question Clarification by dannidin-ga on 09 Dec 2002 00:35 PST ```I think there is some confusion here around the question of: what are M' and N'? It seems to me that what you mean is that M' is the complement of M. Is this true? If so, here is the answer to your question: 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') Let me know if this is what you wanted and I will repost this as an answer. Regards, dannidin``` Clarification of Question by mms03-ga on 09 Dec 2002 00:55 PST ```Dannidin, Your answer is precisely what i was looking for. Sorry for bad symbols. Yes, M' and N' are the complements of M and N```
 Subject: Re: Probability Answered By: dannidin-ga on 09 Dec 2002 01:05 PST Rated:
 ```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')```
 mms03-ga rated this answer: and gave an additional tip of: \$1.00 ```Awesome answer. It was very easy to understand the answer to the problem, with all the steps shown accurately.```

 `Isn't this impossible for the case of M' = N'?`
 ```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.```