Prove that the probability of occurrence of all of a given set of events A(1) through A(n) is equal to the sum of the individual probabilities minus the sum of the probabilities of all pairs of events, A(i) OR A(j) plus the sum of all triples of events, A(i) OR A(j) OR A(k), ..., plus (-1)^(n-1) times the n-tuple A(i) OR ... OR A(n).
Prove for the specific cases of n = 3 and n = 10, and the general case.
We are to prove the required result for events which may not be in general mutually exclusive. Consider any two events A and B. Then the events (A - AB), AB and (B - AB) are pairwise mutually exclusive and we have:
The Sum or Union of two sets A and B is denoted by (A + B) or (A U B) and is defined to be the set of all elements belonging to either A or B or both and the Product or Intersection of two sets A and B is denoted by 'AB' is deifined to be the set of all elements belonging to both A and B.
A = (A - AB) + AB and B = (B - AB) + AB and AB = (A - AB) + AB + (B - AB)
Then, P(A) = P(A - AB) + P(AB) and P(B) = P(B - AB) + P(AB) and P(AB) = P(A - AB) + P(AB) + P(B - AB)
From the above equations, we have:
P(A + B) = P(A) + P(B) - P(AB)
For three events A, B and C, we have:
P(A + B + C) = P(A) + P(B + C) - P{A(B + C)}
[Using the result for the two events A and (B + C) as proved above].
P(A + B + C) = P(A) + P(B + C) - P{A(B + C)}
or, P(A + B + C) = P(A) + P(B) + P(C) - P(BC) - P(AB + AC)
or, P(A + B + C) = P(A) + P(B) + P(C) - P(BC) - [P(AB) + P(AC) - P(AB.AC)]
Using the result for the two events AB and AC and noting that AB.AC = AABC = ABC, we have:
or, P(A + B + C) = P(A) + P(B) + P(C) - P(BC) - P(AB) - P(AC) + P(ABC)
Generalizing for 'n' events, we have the desired result.
Complete Solution
