Help with a problem...

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2 total messages Started by ytakahashi Sun, 11 Oct 1992 21:27

#476

Author: ytakahashi
Date: Sun, 11 Oct 1992 21:27

29 lines
1299 bytes

Perhaps someone can help me out...

 
	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?
 
 
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)
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)
 
Is P(AUBUC) actually what I am looking for?
 
Did I do anything illegal (mathematically speaking) here?
 
If this is right, then the probability of the problem being solved
is nothing more than the addition of the three individual probabilities
minus their product.  Right?
 
Any help is greatly appreciated. 

				- Yuki

please followup to this as my mail server has been unreliable.

Re: Help with a problem...

#479

Author: T. Scott Thompson
Date: Mon, 12 Oct 1992 15:49

40 lines
2142 bytes

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

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