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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: Done !!!! | Comment 6 of 11 |
(In reply to Done !!!! by Ravi Raja)

Remember that when using + or mult for events, + is considered OR, while multiplication (either with a symbol or juxtaposition) is considered to be AND. I think you were using that in A=(A-AB) + AB, but changed over in AB = (A-AB) + AB + (B-AB), which, if I'm reading it right should be A+B = (A-AB) + AB + (B-AB), thus taking the combined A+B area on the Venn diagram and breaking it into its 3 pieces.

But ultimately P(A+B+C) = P(A) + P(B) + P(C) - P(BC) - P(AB) - P(AC) + P(ABC), when interpreted as + = OR and implied multiplication = AND, is an already better known theorem. What is sought is P(ABC)= P(A) + P(B) + P(C) - P(B+C) - P(A+B) - P(A+C) + P(A+B+C) and its generalization.
 Posted by Charlie on 2003-03-15 06:58:09

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