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)^(n1) times the ntuple A(i) OR ... OR A(n).
Prove for the specific cases of n = 3 and n = 10, and the general case.
(In reply to
Complete Solution by Ravi Raja)
Remembering that indeed as you say, "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," the event equation "AB = (A  AB) + AB + (B  AB)" still is wrong in having AB on the left side rather than A+B.
Otherwise your logic is correct in solving P(A + B + C) = P(A) + P(B) + P(C)  P(BC)  P(AB)  P(AC) + P(ABC).
However the goal was to prove P(ABC) = P(A) + P(B) + P(C)  P(B+C)  P(A+B)  P(A+C) + P(A+B+C), and to show indeed how it generalizes to all n.

Posted by Charlie
on 20030317 17:25:47 