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- Newsgroups: sci.math.stat
- Path: sparky!uunet!caen!spool.mu.edu!umn.edu!thompson
- From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
- Subject: Re: Help with a problem...
- Message-ID: <thompson.718904976@daphne.socsci.umn.edu>
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- Reply-To: thompson@atlas.socsci.umn.edu
- Organization: Economics Department, University of Minnesota
- References: <1992Oct11.172756.1@eagle.wesleyan.edu>
- Date: Mon, 12 Oct 1992 15:49:36 GMT
- Lines: 40
-
- 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
-