# Probability of unions/intersections

## Main Question or Discussion Point

Is it true that

Pr( ∪_(n from m to k) ((A_n) ∩ ((A_(n+1))^c)) )
= Pr( ∪_(n from m to k) (A_n) ) - Pr( (A_k) ∩ (A_k+1) )

where A_1, A_2, ... is any sequence of sets.

Well, for the (k=m+1) case I am convinced since I can see they are equal after expanding both sides out, so for example I can see that
Pr((A∩(B^c))∪(B∩(C^c))) = Pr(A∪B) - Pr(B∩C)

but I can't manage to do the same for the (k>m) case in general, so overall I'm not convinced.

Related Set Theory, Logic, Probability, Statistics News on Phys.org
I'm sorry I can't make much sense out of the formula, but assuming its correct, try induction. Assume the proposition is true for k <= m, and add one more term to it and use the truth of the m previous propositions to prove it for (m+1). Will require some grouping and basic properties of sets and cardinalities under union and intersections.

That (induction) is exactly what I've been attempting to use to convince myself that it is true. I've got the base step (k=m+1) which was (for me) expand-able to see that both sides are equal.
However I couldn't get through the inductive step. Perhaps it is false then? Or it could also just mean that I got totally lost within the messy algebra?

HallsofIvy
Homework Helper
You have a "special symbol" right at the beginning of that formula that will not show up on my (or Maverick280857's) browser.

Pr( ∪_(n from m to k) ((A_n) ∩ ((A_(n+1))^c)) )
= Pr( ∪_(n from m to k) (A_n) ) - Pr( (A_k) ∩ (A_k+1) )

or

the probability of [ the union (where n goes from m to k) of [ A_n intersect (A_(n+1) compliment) ] ]

is equal to

the probability of [ the union (where n goes from m to k) of A_n ]
minus
the probability of [ A_k intersect A_(k+1) ]