Triarius Fidelis
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Domine, exaudi vocem meam


« on: September 27, 2007, 10:31:24 pm » 

I read that the probability that at least one of events A and B will occur in a trial is P(A + B) = P(A) + P(B)  P(AB), where A and B are not mutually exclusive events. That makes sense, because one would count the probabilities of A and B together, and then ignore event AB, because that counts instances of A and B twice, in the same way that the probability that at least one die of two will turn up six is 11/36, not 1/3.
That made sense intuitively, but I couldn't quite follow the proof because it mentioned things I never saw before. I noticed however, that the following relation also appears to be true (A_{C} and B_{C} are complementary events because latex2html is a dick.)
P(A_{C}B_{C}) + P(AB) + P(A_{C}B) + P(AB_{C}) = 1 P(AB) + P(A_{C}B) + P(AB_{C}) = 1  P(A_{C}B_{C})
and so
P(A + B) = 1  P(A_{C}B_{C})
I did it on some practice examples and it made sense, but what do I know...


« Last Edit: October 02, 2007, 05:01:33 am by hanumizzle »

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carsten
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« Reply #1 on: September 28, 2007, 02:57:38 am » 

The solution is 42
Carsten
... and thanx for the fish



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exeterdad
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« Reply #2 on: September 28, 2007, 04:09:32 am » 

P(A_{C}B_{C}) + P(AB) + P(A_{C}B) + P(A_{C}B) = 1 P(AB) + P(A_{C}B) + P(A_{C}B) = 1  P(A_{C}B_{C})
and so
P(A + B) = 1  P(A_{C}B_{C})
Whaddid he say?? One of these days that brain of yours is going to blow up Hanu.



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rbistolfi
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« Reply #3 on: September 28, 2007, 07:23:05 am » 

God, I missed you



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M0Elnx


« Reply #4 on: September 28, 2007, 07:27:06 am » 

WoW..... My brains just overheated from just reading and trying to make sense of this thread....
I better leave it alone



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saulgoode


« Reply #5 on: September 28, 2007, 08:39:21 am » 

Shouldn't the starting point for the complementary case be stated:
P(A_{C}B_{C}) + P(AB) + P(A_{C}B) + P(AB_{C}) = 1


« Last Edit: September 28, 2007, 08:46:40 am by saulgoode »

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Vanger


« Reply #6 on: September 28, 2007, 08:49:26 am » 

Use the common graphical representation via circles and everything should be clear.



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MikeCindi
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« Reply #7 on: September 28, 2007, 10:01:44 am » 

Shouldn't the starting point for the complementary case be stated:
P(A_{C}B_{C}) + P(AB) + P(A_{C}B) + P(AB_{C}) = 1
That's correct...and I believe that the way Hanu worked his query is using this starting point versus his. Thus just a typo on his part. I'm not sure that I'm the one to comment on whether he got it right though...


« Last Edit: September 28, 2007, 10:04:34 am by mikecindi »

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Triarius Fidelis
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Domine, exaudi vocem meam


« Reply #8 on: October 02, 2007, 05:02:14 am » 

Shouldn't the starting point for the complementary case be stated:
P(A_{C}B_{C}) + P(AB) + P(A_{C}B) + P(AB_{C}) = 1
Yes...



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