p(A1 U A2 U ...UAk) = p(1) + p(2) + ... + p(k)

p(A1 U A2 U ...UAk) = p(1) + p(2) + ... + p(k)




Let be k a finite number of trials such that in each one there are the probability p of success, and, therefore the probability 1-p of failure.

By the Binomial Theorem

( (1-p )+ p)) ^ k =
________sum (kCj* (1-p)^j )______ 0<= j <=k

Therefore

_1 = (1-p) ^k + sum (kCj* (1-p)^j )____ 1<= j <=k

The sum is
____p(at least 1 success) = p(A1 U A2 U ...UAk)

Finally

Let be p(j) the probability to occur exactly j successes

p(A1 U A2 U...UAk) = p(1) + p(2) + ... + p(k)

which is the formula that we intended to derive.


The result is rather banal: the probability to observe at least 1 success is the sum of the probability to have 1 success plus the probability to have 2 successes ... plus the probability to have only successes, no failures, at the k trials.
It was supposed that A1, A2, ... , Ak are independent mutually exclusive events (as are coins tossing or flipping). If not we have

___ p(A1 U A2 U …UAk) < p(1) + p(2) + …+ p(k)

instead (Boole´s Inequality).



____licas (Luis A. Afonso)
.

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