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Probability of unions/intersections

  1. Mar 14, 2008 #1
    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. jcsd
  3. Mar 15, 2008 #2
    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.
  4. Mar 15, 2008 #3
    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?
  5. Mar 15, 2008 #4


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    You have a "special symbol" right at the beginning of that formula that will not show up on my (or Maverick280857's) browser.
  6. Mar 15, 2008 #5
    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) )


    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 ]
    the probability of [ A_k intersect A_(k+1) ]
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