VectorLinux
The Vectorian Lounge => The Lounge => Topic started by: Triarius Fidelis on September 27, 2007, 11:31:24 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...

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

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.

God, I missed you :D

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

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

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

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...

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... :[