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Re: Conditional Probability Question
Posted:
Sep 26, 2001 12:27 AM
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aaron@avalon.pascal-central.com (Aaron Davies) writes:
> Randy Poe <rpoe@atl.lmco.com> wrote:
> > Interpretation #2: Given that I have already pulled out
> > (i-1) parts and found them to be good, what is
> > the probability that the next part is bad? In that
> > case, the answer is 1/(100 - (i-1)) = 1/(101-i).
>
> Yes, that's the basic line of thinking the students are using. The
> problem is that the professor presented math along the lines of what
> Timothy E. Vaughan posted, quoted below.
>
> > The probability that you inspect a second piece is therefore
> > 99/100. The probability that you inspect it and find it to be
> > defective is (99/100)*(1/99) = 1/100.
>
> This is supposed to reflect the second interpretation. It's that that I
> don't get.
Notice carefully:
"The probability that you inspect it AND find it to be defective..."
The probability that you inspect it is (99/100), the probability that
it is defective given that you are inspecting it is (1/99).
Contrast that with
"Given that I have already pulled out (i-1) parts and found them to be
good..."
In this case the probability that you inspect the next part is 1.
But the question asked, what is the probability that the defective
part is found at the i-th inspection. It did not say "given that the
first i-1 were not defective".
Kevin.
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