
Re: Conditional Probability Question
Posted:
Sep 27, 2001 2:00 AM


>Subject: Re: Conditional Probability Question >From: Virgil vmhjr2@home.com >Date: 9/26/01 11:31 PM Central Daylight Time >Messageid: <vmhjr246A44B.22411426092001@news> > >In article <20010926231037.07418.00001024@mbcu.aol.com>, > mensanator@aol.com (Mensanator) wrote: > >> >Actually, it doesn't. I was mistaken before (blush!). Let's give a >> >formal argument to find P(i) = prob. that the ith part is found to be >> >defective. P(1)=1/100. Now i=2 occurs if the first is nondefective >> >(prob. = 99/100) while the second one is defective (prob = 1/99, since >> >only 1 of the remainig 99 is bad). Thus P(2) = (99/100)(1/99)= 1/100. >> >Similarly, P(3)=(99/100)(98/99)(1/98) = 1/100, etc. >> >> Yeah, I had worked that out but was afraid I had overlooked something. >> So, Virgil's comments not withstanding, I don't have to alter my web page. > >I repeat that the way you should caluculate probability depends on >what you know ( or when you do the calculation) . > >If you are calculating the probability that the i'th part is >defective BEFORE inspecting any parts, AS YOU ARE DOING, the >probability is 1/100 AS YOU FOUND, regardless of how the inspections >are to be done.
Ok.
> >Once you have inspected any parts, the probabilities with respect to >the remaining parts AT THAT TIME are no longer the same as they were >before any inspections. They are now conditional probabilities, >conditional upon finding what was actually found.
I've been saying the same thing. When I say that for i=4 the probability is
1/97 * 97/98 * 98/99 * 99/100
the 1/97 term is the conditional probability that the ith inspection discovers the defect. The other terms are the conditional probabilities that the previous inspections failed to find the defect.
> >Note: If the pieces are to be inspected in sequence: > > >P(2) = P( (not 1) and 2) > = P( not 1) * P( 2 given (not 1) ) > > >Before any inspections, P (not 1 ) = 99/100 >and P( 2 given ( not 1) ) = 1/99, so P(2) = 1/100, as expected
Ok.
> > >Once the first has been examined and found to be good, >P(1) = 0 and P(not 1) = 1 and >P(2) = P(not1)*P(2 given not 1) = 1*(1/99) = 1/99 > >After the first and second have been examined, > P(2) = 1 if 2 was found to be defective or > P(2) = 0 if 2 was found to be good.
You are confusing the "outcome" of an event with it's "probability". An outcome does not change the probability. It was 1/100 before you inspected it and it's probability was (note past tense) 1/100 after you inspected it.

