# is my proof correct

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### #1dionys

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Posted 17 October 2004 - 07:10 AM

Hi...Can you plz check if my proof is correct?

Exercise:
A1,A2,.....An are independently events.
Prove that :
P(A1[union]A2[union]...[union]An) = 1-Πi[element-of]I(1-P(Ai))

note for this (Πi[element-of]I(1-P(Ai))
I={1,2,....n)
P([intersect]Ai)= Π P(Ai)
for 3 events A1,A2,A3
means: P(A1[intersect]A2)=P(A1)*P(A2)
P(A2[intersect]A3)=P(A2)*P(A3)
P(A2[intersect]A3)=P(A2)*P(A3)
P(A1[intersect]A2[intersect]A3)=P(A1)* P(A2) * P(A3)

Now my proof:
We know that P([intersect]Ai)= Π P(Ai)
if A1,A2,...,An are independent then and the complements
are independent

P([intersect]Ai)complement = Π P(Aicomplement)
P([union](Ai compl) ) = Π(1-P(Ai))
1-P([union]Ai)= Π(1-P(Ai))
-P([union]Ai)=-1+Π(1-P(Ai))
Finally ... we got our proof
P([union]Ai)=1-Πi[element-of]I(1-P(Ai))
Is it correct?

### #2Per Vognsen

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Posted 18 October 2004 - 01:35 AM

dionys said:

Hi...Can you plz check if my proof is correct?

Exercise:
A1,A2,.....An are independently events.
Prove that :
P(A1[union]A2[union]...[union]An) = 1-Πi[element-of]I(1-P(Ai))

note for this (Πi[element-of]I(1-P(Ai))
I={1,2,....n)
P([intersect]Ai)= Π P(Ai)
for 3 events A1,A2,A3
means: P(A1[intersect]A2)=P(A1)*P(A2)
P(A2[intersect]A3)=P(A2)*P(A3)
P(A2[intersect]A3)=P(A2)*P(A3)
P(A1[intersect]A2[intersect]A3)=P(A1)* P(A2) * P(A3)

Now my proof:
We know that P([intersect]Ai)= Π P(Ai)
if A1,A2,...,An are independent then and the complements
are independent

P([intersect]Ai)complement = Π P(Aicomplement)
P([union](Ai compl) ) = Π(1-P(Ai))
1-P([union]Ai)= Π(1-P(Ai))
-P([union]Ai)=-1+Π(1-P(Ai))
Finally ... we got our proof
P([union]Ai)=1-Πi[element-of]I(1-P(Ai))
Is it correct?

Dude, your notation is ghetto as fuck. Here's a shorter proof:

Let A = A_1 union ... union A_n. For a set X, let c(X) denote its complement.

P(A) = P(c(c(A)) = P(c(c(A_1) intersect ... intersect c(A_n))) = 1 - P(c(A_1) intersect ... intersect c(A_n)))

I used De Morgan's law for the second equality. The complements of independent events are independent so the above equals

1 - prod c(A_i) = 1 - prod(i=1..n) (1 - P(A_i))

Done.

### #3Nick

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Posted 19 October 2004 - 01:57 PM

dionys, you have five posts in total at this moment, and they are all homework. You seem to be lucky that a lot of people want to help you, but this is not where these forums are for. I'm not an administrator or anything, but I just want to warn you. Besides, if you can't solve your homework yourself, I also doubt that you will be succesful at your exams, and even your further professional life. Aim high, but pull the bow yourself.

### #4tgfx

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Posted 02 November 2004 - 08:43 PM

That’s going to encourage him…nice going Nick...lol
Can't you guys be nice...

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### #5Ed Mack

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Posted 02 November 2004 - 09:34 PM

It's kinda the hard truth though..

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Posted 03 November 2004 - 07:31 AM

I think it is *meant* to discourage him. Not all activities should be encouraged.
Jesse Coyle

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