Question

Addition theorem of probability for 3 events proof

Answer 1

Let PrPr be a probability measure on an event space ΣΣ.

Let A,B∈ΣA,B∈Σ.


Then:

Pr(A∪B)=Pr(A)+Pr(B)−Pr(A∩B)Pr(A∪B)=Pr(A)+Pr(B)−Pr(A∩B)

That is, the probability of either eventoccurring equals the sum of their individual probabilities less the probability of them both occurring.


This is known as the addition law of probability.


Proof 1

By definition, a probability measure is a measure.

Hence, again by definition, it is a countably additive function.

By Measure is Finitely Additive Function, we have that PrPr is an additive function.

So Additive Function is Strongly Additive can be applied directly.

■◼


Proof 2

From Set Difference and Intersection form Partition:

AA is the union of the two disjoint sets A∖BA∖B and A∩BA∩BBB is the union of the two disjoint sets B∖AB∖A and A∩BA∩B.


So, by the definition of probability measure:

Pr(A)=Pr(A∖B)+Pr(A∩B)Pr(A)=Pr(A∖B)+Pr(A∩B)Pr(B)=Pr(B∖A)+Pr(A∩B)Pr(B)=Pr(B∖A)+Pr(A∩B)


From Set Difference Disjoint with Reverse:

(A∖B)∩(B∖A)=∅(A∖B)∩(B∖A)=∅


Hence:

Pr(A)+Pr(B)Pr(A)+Pr(B)==Pr(A∖B)+2Pr(A∩B)+Pr(B∖A)Pr(A∖B)+2Pr(A∩B)+Pr(B∖A)==Pr((A∖B)∪(A∩B)∪(B∖A))+Pr(A∩B)Pr((A∖B)∪(A∩B)∪(B∖A))+Pr(A∩B) Set Difference and Intersection form Partition: Corollary==Pr(A∪B)+Pr(A∩B)Pr(A∪B)+Pr(A∩B)

Hence the result.