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

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28 Jul 2003, 13:46

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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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Ride em cowboy

Last edited by Curly05 on 28 Jul 2003, 20:14, edited 1 time in total.

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

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28 Jul 2003, 14:45

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:

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

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? Thanks- please explain more Akami
Victor
_________________

Ride em cowboy

Last edited by Curly05 on 29 Jul 2003, 05:10, edited 1 time in total.

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.
_________________

Best,

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