Probabilities and independence

[dionys]

Hi...Can you please 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?


And one more...
but i don't know how to prove this:
A,B,C are independent
We must prove that A and B[union]C are independent too
...?

 

[Galileo]

I couldn't follow some steps in your derivation (then again, it's pretty early here and I`m half asleep).

(Try using LateX, it makes your notation look more sexy. Click on the formulas to see the code.)

I would do the following:
$$P\left(\bigcup_{i=1}^n A_i\right)=1-P\left((\bigcup_{i=1}^n A_i)^c\right)=1-P\left(\bigcap_{i=1}^n A_i^c\right)=1-\prod_{i=1}^nP\left(A_i^c\right)=1-\prod_{i=1}^n(1-P\left(A_i)\right)$$

 

[wais]

Hi there,

Your proof for the first exercise is correct! Good job.

For the second exercise, you can use the definition of independence to prove it. Recall that two events A and B are independent if and only if P(A[intersect]B) = P(A)*P(B). So, to prove that A and B[union]C are independent, we need to show that P(A[intersect](B[union]C)) = P(A)*P(B[union]C).

To do this, we can use the fact that P(A[intersect](B[union]C)) = P((A[intersect]B)[union](A[intersect]C)), and then use the definition of independence for A[intersect]B and A[intersect]C. Can you take it from here?

Let me know if you have any other questions or need further clarification. Good luck!