Probability of unions/intersections

In summary, the conversation is discussing the equality of two probability equations involving sequences of sets. The first equation expands to show equality for the (k=m+1) case, but the speakers are unable to prove the same for the (k>m) case using induction. One speaker suggests using grouping and basic properties of sets to prove it, but another speaker points out a special symbol in the formula that may cause issues.
  • #1
e12514
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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.
 
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  • #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.
 
  • #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?
 
  • #4
You have a "special symbol" right at the beginning of that formula that will not show up on my (or Maverick280857's) browser.
 
  • #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) )



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) ]
 

1. What is the probability of a union of two events?

The probability of the union of two events A and B is calculated by adding the individual probabilities of the two events and subtracting the probability of their intersection. It can be expressed as P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

2. How can we find the probability of an intersection of two events?

The probability of the intersection of two events A and B is calculated by multiplying the individual probabilities of the two events. It can be expressed as P(A ∩ B) = P(A) * P(B).

3. What is the difference between a union and an intersection in terms of probability?

A union refers to the combination of two events, while an intersection refers to the occurrence of both events together. In terms of probability, a union is calculated by adding probabilities, while an intersection is calculated by multiplying probabilities.

4. How does the probability of a union change with more than two events?

The probability of a union with more than two events can be calculated by adding the individual probabilities of all the events and then subtracting the probabilities of all possible intersections between the events. It can be expressed as P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C).

5. What is the probability of an intersection if the events are independent?

If two events are independent, the probability of their intersection is calculated by multiplying their individual probabilities. In other words, the probability of one event does not affect the probability of the other event. It can be expressed as P(A ∩ B) = P(A) * P(B).

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