If  M and N are independent events,  P(M,N) = P(M)P(N)   then prove
that M' and N' are independent.

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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

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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"

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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

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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

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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

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')

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.