
Re: Conditional Probability Question
Posted:
Sep 26, 2001 9:40 AM


>Subject: Re: Conditional Probability Question >From: Virgil vmhjr2@home.com >Date: 9/25/01 11:23 PM Central Daylight Time >Messageid: <vmhjr2DECB08.22331625092001@news1.denver1.co.home.com> > >In article <20010925234846.05795.00000711@mbmf.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 >> >Messageid: <3BB10091.36487FAB@engmail.uwaterloo.ca> >> > >> >There is a difference between examining previous parts and not examining >> >them. If you take out (i1) 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 setaside 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 i1. 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 i1 inspections, so it doesn't matter whether you inspect them or not.

