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Author Topic: theoretical probability question for mathematicians  (Read 2169 times)
Triarius Fidelis
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« 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 (AC and BC are complementary events because latex2html is a dick.)

P(ACBC) + P(AB) + P(ACB) + P(ABC) = 1
P(AB) + P(ACB) + P(ABC) = 1 - P(ACBC)

and so

P(A + B) = 1 - P(ACBC)

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 » Logged

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

The solution is 42

Carsten

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


P(ACBC) + P(AB) + P(ACB) + P(ACB) = 1
P(AB) + P(ACB) + P(ACB) = 1 - P(ACBC)

and so

P(A + B) = 1 - P(ACBC)

Whaddid he say??  Huh  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  Cheesy
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M0E-lnx
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« 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
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« Reply #5 on: September 28, 2007, 08:39:21 am »

Shouldn't the starting point for the complementary case be stated:

P(ACBC) + P(AB) + P(ACB) + P(ABC) = 1

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

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Vanger
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« 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(ACBC) + P(AB) + P(ACB) + P(ABC) = 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 » Logged

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

Shouldn't the starting point for the complementary case be stated:

P(ACBC) + P(AB) + P(ACB) + P(ABC) = 1



Yes...  Embarrassed
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