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Topic: interesting probability problem
Replies: 9   Last Post: Oct 17, 2007 3:20 AM

 Messages: [ Previous | Next ]
 Gus Gassmann Posts: 181 Registered: 12/7/04
Re: interesting probability problem
Posted: Oct 16, 2007 8:47 AM

On Oct 16, 11:25 am, Andersen <andersen_...@hotmail.com> wrote:
> Bertsekas has the following exercise in his probability book:
>
> "
> Consider a statement whose truth is unknown. If we see many examples
> that are compatible with it, we are tempted to view the statement as
> more probable. Such reasoning is often referred to as inductive
> inference (in a philosophical, rather than mathematical sense). Consider
> now the statement that ?all cows are white.? An equivalent statement is
> that ?everything that is not white is not a cow.? We then observe
> several black crows. Our observations are clearly compatible with the
> statement, but do they make the hypothesis ?all cows are white? more likely?
>
> To analyze such a situation, we consider a probabilistic model. Let us
> assume that there are two possible states of the world, which we model
> as complementary events:
>
> A : all cows are white,
> C(A) : 50% of all cows are white. [C(A) is the complement event of A].
>
> Let p be the prior probability P(A) that all cows are white. We make an
> observation of a cow or a crow, with probability q and 1?q,
> respectively, independently of whether event A occurs or not. Assume
> that 0 < p < 1, 0 < q < 1, and that all crows are black.
>
> (a) Given the event B = {a black crow was observed}, what is P(A|B)?
> (b) Given the event C = {a white cow was observed}, what is P(A|C)?
> "
>
> ---
> Solutions to a) is p, and b) is 2p/(1-p). From this he draws the
> conclusion that a) does not affect the hypothesis A, while b)
> strengthens it. Is this reasoning correct? I mean event B should have
> the same effect as event C, as B supports A, since A is equivalent to
> "everything that is not white is not a cow". What is wrong in my reasoning?

Two quick observations. First, the solution to b) must be wrong,
because
the expression is > 1 if p is > 1/3. Second, the states of nature A
and C(A)
are very strangely chosen, as there is a lot of middle ground. C(A)
should be
"There exists a cow that is not white". If you then try to apply
Bayes' theorem,
you find that P(C|comp(A)) is no longer 1/2.

Date Subject Author
10/16/07 Andersen
10/16/07 Gus Gassmann
10/16/07 Andersen
10/16/07 matt271829-news@yahoo.co.uk
10/16/07 Andersen
10/16/07 briggs@encompasserve.org
10/16/07 matt271829-news@yahoo.co.uk
10/17/07 Andersen
10/17/07 Andersen
10/16/07 Robert Israel