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Re: Conditional Probability Question
Posted:
Sep 28, 2001 12:32 AM
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>Subject: Re: Conditional Probability Question
>From: aaron@avalon.pascal-central.com (Aaron Davies)
>Date: 9/27/01 7:07 PM Central Daylight Time
>Message-id: <1f0eso5.1pgqnqui96pf4N%aaron@avalon.pascal-central.com>
>
>Mensanator <mensanator@aol.com> wrote:
>
>> No it isn't. If there were 5 parts with the defect at i=3, you get
>>
>> p(1) = 0
>> p(2) = 0
>> p(3) = 1/(5-3+1) = 1/3
>> p(4) = 1/(5-4+1) = 1/2
>> p(5) = 1/(5-5+1) = 1
>>
>> This sums to 11/6.
>
>This is because you're adding probabilities of different events.
Actually, it is because I misinterpreted Virgil's equations. Disregard the
above.
>At any
>given time, the probabilities sum to 1. Right before test 3, the probs
>are 0, 0, 1/3, 1/3, 1/3. Right before test 4, the probs are 0, 0, 0,
>1/2, 1/2. Right before test 5, they're 0, 0, 0, 0, 1.
Right. The probability of test 5 succeeding is 1. Provided you get to test 5.
To get to test 5, test 4 must fail. To get to test 4, test 3 must fail. To get
to
test 3, test 2 must fail. To get to test 2, test 1 must fail. You won't even do
test 5 unless all previous tests fail.
The probability of failure is (1-p) where p is the probability of success.
For test 1, p = 1/5 so (1-p) = 4/5
For test 2, p = 1/4 so (1-p) = 3/4
For test 3, p = 1/3 so (1-p) = 2/3
For test 4, p = 1/2 so (1-p) = 1/2
For test 5, p = 1/1 so (1-p) = 0
So if we combine the probability that test 5 will succeed (1/1) with the
probabilities that each previous test fails (which is done by multiplying them
all together) you get
1/1 * 1/2 * 2/3 * 3/4 * 4/5 = 1/5
and you'll find that the overall probability of each test is 1/5.
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