>Subject: Re: Conditional Probability Question >From: Virgil firstname.lastname@example.org >Date: 9/25/01 11:23 PM Central Daylight Time >Message-id: <vmhjr2-DECB08.email@example.com> > >In article <firstname.lastname@example.org>, > email@example.com (Mensanator) wrote: > >> >Subject: Re: Conditional Probability Question >> >From: Ray Vickson firstname.lastname@example.org >> >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.