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# P(1 defect) = 2 * (999/1000) * (1/1000) = There is

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P(1 defect) = 2 * (999/1000) * (1/1000) = There is [#permalink]

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28 Jul 2003, 14:46
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P(1 defect) = 2 * (999/1000) * (1/1000) =

There is now way this eq up here is right.

. It is about independent events.

A manufacturer makes fastners and sells in packages of two. Suppose the condition of a fastener is independent of the other fastener in the package and in the past the fraction of defective light bulbs is 1/1000. What is the probability that a randomly chosen package has a.) 0 defects b.) Exactly one defect c.) Exactly two defects
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Last edited by Curly05 on 28 Jul 2003, 21:14, edited 1 time in total.

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Re: PS: Easy Prob, but author has error? [#permalink]

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28 Jul 2003, 15:45
Curly05 wrote:
P(1 defect) = 2 * (999/1000) * (1/1000) =

There is now way this eq up here is right.

. It is about independent events.

A manufacturer makes fastners and sells in packages of two. Suppose the condition of a fastener is independent of the other fastener in the package and in the past the fraction of defective light bulbs is 1/1000. What is the probability that a randomly chosen package has a.) 0 defects b.) Exactly one defect c.) Exactly two defects

Let's say you examine them one by one. The probability that the first is defective is 1/1000 AND the probability that the second is not defective is 999/1000 so the probability of this event is (1/1000)(999/1000). But the defective fastener could be in either the first or second position (i.e., there are two ways that there could be just one defective fastener) so we need to multiply this probability by 2, hence:

Pr(exactly 1 defect) = 2*(1/1000)(999/1000).
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Eternal Intern
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28 Jul 2003, 18:27
So P( 2 defects) , the first defect is reserved for the first slot, and the second defect is reserved for the second slot, thus only one outcome for each slot.

P( 1 defect) , there are two outcomes for defective slot, right- the first or the second. and the prob is etc.

Let's change it around and let's say they want to find the probablilty of defect-free slots.
P( 1 defect-free), Are there are two outcomes for the defect-free slot,right?
Victor
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Last edited by Curly05 on 29 Jul 2003, 06:10, edited 1 time in total.

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29 Jul 2003, 00:55
P(a bulb is defect)=0.001
P(a bulb is OK)=0.999

no defects) 0.999*0.999
exactly 1 defect) 0.999*0.001+0.001*0.999=2*0.999*0.001
two defects) 0.001*0.001

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Eternal Intern
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29 Jul 2003, 10:02
Stolyar, is your exactly one defect showing that there are four drawings?

VT
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29 Jul 2003, 10:16
Curly05 wrote:
Stolyar, is your exactly one defect showing that there are four drawings?

VT

There are only TWO drawings. Since we are specifically looking at ONE defect, the fact that the defective fastener is picked first is exactly the same as saying the non-defective fastener is picked second. So you can describe the events from either the point of reference using defective fasteners, or non-defective fasteners, but not both.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: Hey Stolyar   [#permalink] 29 Jul 2003, 10:16
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