# Probability of unions/intersections

1. Mar 14, 2008

### e12514

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.

2. Mar 15, 2008

### maverick280857

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.

3. Mar 15, 2008

### e12514

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?

4. Mar 15, 2008

### HallsofIvy

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

5. Mar 15, 2008

### e12514

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) ]