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


Kevin E. Thorpe wrote:
Anon. wrote:
\"Luis A. Afonso\" wrote:

...
> It was supposed that
> A1, A2, ... , Ak are independent mutually exclusive events (as are
> coins tossing or flipping).
>
Surely it is possible to have independent mutually exclusive events!
Well, excepting the trivial case that some have probability 0.

Bob

I hope you meant to say it is impossible to have independent
mutually exlusive events.

Isn't it amazing that Afonso made a blunder in the high-school level
equality in probability of union of events; and then Bob O'Hara made
another high-school level error while failing to correct Afonso's
error!

A and B are indepenent iff P(AB) = P(A)P(B).
If A and B are mutually exclusive then P(AB) = 0.
So if A and B are independent and
P(A) > 0 and P(B) > 0 then P(AB) > 0 and
therefore not mutually exclusive.

As I always tell my students that the concepts of "mutually
exclusiveness"
and "independence" of events are themselves mutually exclusive --
because you can NEVER have both!

As for the IDENTITY and implied inequalities, of any arbitrary number
of events on the union, it is the generalization of

P(A1 U A2 U A3) = P(A1) + P(A2) + P(A3) - [P(A1 and A2) +
P(A1 and A3) + P(A2 and A3)] + P(A1 and A2 and A3).

-- Reef Fish Bob.





--
Kevin E. Thorpe
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto

.

Relevant Pages