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 Probability of All of a Set (Posted on 2003-03-13)
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.

 See The Solution Submitted by Charlie Rating: 3.2500 (4 votes)

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 re: Complete Solution Comment 11 of 11 |
(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 2003-03-17 17:25:47

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