@dionys

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 ) = 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?[snapback]12928[/snapback]

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.

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 complementsare independent

P([intersect]Ai)complement = Π P(Aicomplement)

P(union ) = 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?