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

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 mensanator Posts: 5,039 Registered: 12/6/04
Re: Conditional Probability Question
Posted: Sep 28, 2001 12:00 AM

>Subject: Re: Conditional Probability Question
>From: Virgil vmhjr2@home.com
>Date: 9/27/01 3:22 PM Central Daylight Time
>Message-id: <vmhjr2-EDD1B1.14315727092001@news1.denver1.co.home.com>
>
>In article <20010927020033.25553.00000712@mb-cr.aol.com>,
> mensanator@aol.com (Mensanator) wrote:
>

>> 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.

>
>
>This calculation is only valid before any inspection. After the
>inspection of the first part, the "probability" of that part being
>good no longer exists, since the determination has already occured.

Ok.

>
>Once an experiment has been completed and observed, any of its
>observable outcomes has "probability" either equal to 0 or be equal
>to 1, with no intermediate possibilities.

I see that now. I misinterpreted your equations before.

>
>If Schreodinger's cat has been seen to be dead, then you can no
>longer say that the probability of its being dead is 1/2.

But I still say

"What is the probability that the i-th part you examine will be defective?"

means to calculate the probability before inspection begins. If playing
Russian Roulette, we can ask

"What is the probability that the i-th player will shoot himself?"

In this case, we "inspect" the chamber for the presence of a bullet by
pulling the trigger. Now you could certainly interpret that question to mean
player 6's probability of finding the bullet is 1/1 (which is what it would be
if all previous players clicked on an empty chamber), but that gives the
impression
that the game is unfair. But this is false. The game _is_ fair. By fair we mean
that each player has the same probability of finding the bullet: 1/6.
Therefore,
the correct interpretation is to calculate probability before inspection
begins.

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