🚀 go-pugleaf

ytakahashi@eagle.wesleyan.edu writes:

>	Three people work to solve a problem independantly with differing
>chances of solving it.  Given these probabilities, what is the probability
>that the problem is solved?

This looks a lot like homework so I won't give a complete answer:

>Since they are independent, P(AB) = P(A)*P(B)  by definition
>I extended this to... P(ABC) = P(A)*P(B)*P(C)
>	(I'm not sure if that's legal to do)

Yes, this is legal.  It is the definition of independence for 3 events.

Aside: P(AB)=P(A)*P(B), P(BC)=P(B)*P(C) and P(AC)=P(A)*P(C) does _not_
imply that P(ABC)=P(A)*P(B)*P(C).  So although you are using the
correct expression for P(ABC), the reasoning that got you there is not
necessarily correct.

>P(AUBUC) = P(A)+P(B)+P(C) - P(ABC)   by definition

>plug in from above yields...
>P(AUBUC) = P(A)+P(B)+P(C) - P(A)*P(B)*P(C)

But this is wrong.  (If you are not convinced, try P(A)=P(B)=P(C)=1/2
in the formula.  Then the right-hand side is 3/2 - 1/8 = 11/8, which
is not a probability.)  You are incorrectly assuming here that

P(AUBUC) = P(A) + P(B) + P(C) - P(ABC)

I will leave it up to you to figure out why this is not correct, and
what correction is needed.

Hint: Write out the 2*2*2 = 8 different outcomes and count how many
times each one is included on the right-hand side of your formula.  A
Venn diagram of the possibilities will help.
--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209