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:
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)
Eliminating P(A -AB) and P(B - AB) from the above equations, we get the general addition rule (for two events, which are not mutually exclusive) which is represented mathematically as follows:
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 following result:
P[(A1) + (A2) + .... + (An)] = P(A1) +
P(A2) + .... + P(An) - P[%
Solution !!!!
