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Re: Conditional Probability Question
Posted:
Sep 26, 2001 9:40 AM
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>Subject: Re: Conditional Probability Question
>From: Virgil vmhjr2@home.com
>Date: 9/25/01 11:23 PM Central Daylight Time
>Message-id: <vmhjr2-DECB08.22331625092001@news1.denver1.co.home.com>
>
>In article <20010925234846.05795.00000711@mb-mf.aol.com>,
> mensanator@aol.com (Mensanator) wrote:
>
>> >Subject: Re: Conditional Probability Question
>> >From: Ray Vickson rvickson@engmail.uwaterloo.ca
>> >Date: 9/25/01 5:09 PM Central Daylight Time
>> >Message-id: <3BB10091.36487FAB@engmail.uwaterloo.ca>
>> >
>> >There is a difference between examining previous parts and not examining
>> >them. If you take out (i-1) parts and put them aside without examining
>> >them, the probability that the i'th part (which IS examined) will be
>> >defective is 1/100. Surprisingly, this is true even if i=100, so you are
>> >drawing from a box that contains only 1 part. However, everything
>> >changes if the set-aside parts are examined instead of ignored.
>>
>> Why does it change?
>
>One's evaluation of probability is generally dependent on one's
>state of knowledge at the time of the evaluation. Each part
>examined, up to finding the defective part, changes one's state of
>knowledge.
I don't see that. What changes is that on every inspection, the probabilty of
finding the part goes from 1/100 to 1/1. But that is _not_ the probability that
the part is _at_ position i. It is the probability of finding it at postion i
_combined_ with the probability that it is _not_ at posititions i-1. And if you
do the correct math, you'll see that this is 1/100 for every i.
And the probability the ith part is defective does not change based on your
knowledge of the previous i-1 inspections, so it doesn't matter whether you
inspect them or not.
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