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Topic: Conditional Probability Question
Replies: 49   Last Post: Oct 18, 2001 2:34 PM

 Messages: [ Previous | Next ]
 mensanator Posts: 5,039 Registered: 12/6/04
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
>Message-id: <vmhjr2-46A44B.22411426092001@news>
>
>In article <20010926231037.07418.00001024@mb-cu.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 non-defective
>> >(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.

Date Subject Author
9/25/01 Aaron Davies
9/25/01 Randy Poe
9/25/01 Aaron Davies
9/26/01 Kevin Foltinek
9/26/01 mensanator
9/27/01 Kevin Foltinek
9/27/01 Aaron Davies
9/27/01 mensanator
9/25/01 Steve Wright
9/25/01 Timothy E. Vaughan
9/26/01 Lionel Barnett
9/25/01 Virgil
9/25/01 Ray Vickson
9/25/01 mensanator
9/26/01 Virgil
9/26/01 mensanator
9/26/01 Virgil
9/27/01 mensanator
9/27/01 The Scarlet Manuka
9/27/01 mensanator
9/27/01 Virgil
9/27/01 Aaron Davies
9/28/01 mensanator
9/26/01 Ray Vickson
9/26/01 mensanator
9/27/01 Virgil
9/27/01 mensanator
9/27/01 mensanator
9/27/01 Virgil
9/28/01 mensanator
9/28/01 Virgil
9/28/01 David Lloyd-Jones
9/28/01 mensanator
9/29/01 David Lloyd-Jones
9/29/01 mensanator
9/29/01 David Lloyd-Jones
9/29/01 mensanator
9/29/01 David Lloyd-Jones
9/29/01 mensanator
9/29/01 Virgil
9/29/01 mensanator
9/29/01 Lynn Kurtz
9/27/01 Aaron Davies
9/27/01 mensanator
9/27/01 Virgil
10/17/01 Mike Mccarty Sr
10/18/01 Virgil
9/27/01 Randy Poe
9/25/01 mensanator