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Problem :

A lot of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non defective bulbs. If 8 bulbs are selected at random, what is the probability that there will be at least one defective bulb.

Method I :

Probability that no bulb will be defective , $P(X =0) $

$$P(0) ={}^8C_0 p^0 q^8 =q^8 = (\frac{9}{10})^8$$

where q probability that a bulb selected is non defective $$\therefore q = 1-p = 1-\frac{1}{10} = \frac{9}{10}$$

and p ( probability of bulb drawn is a defective ) = $$\frac{10}{100} = \frac{1}{10}$$

Now probability that at least one bulb is defective = $$1 -P(0) = 1-(\frac{9}{10})^8$$

Method II :

But I want to find the answer in another way,

Probability of drawing non defective bulb

$$= \frac{90}{100} \times \frac{89}{99} \times \frac{88}{98} \times...........\times \frac{82}{92} ..........(i)$$

Now the probability of drawing at least one bulb will be defective

$$= 1-(i) = 1- \frac{90}{100} \times \frac{89}{99} \times \frac{88}{98} \times...........\times \frac{82}{92}$$

But this is not the correct answer ... please suggest the correction here..... thanks...

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    $\begingroup$ The difference is that the first method includes reposition while the second method does not. The first bulb being good is in either case $\frac{90}{100}=\frac9{10}$, once you picked that bulb the second bulb will either be $\frac{90}{100}=\frac9{10}$ with reposition, or $\frac{89}{99}$ without it. $\endgroup$ – Carlos Eugenio Thompson Pinzón Oct 24 '13 at 16:23
  • $\begingroup$ The problem is, at best, ambiguously phrased. It is not made clear whether you are selecting bulbs from a given lot with replacement (in which case method I is required) or without replacement (in which case method II is required). Yet a third (I think less likely) interpretation is that the $10\%$ failure rate represents an overall failure rate, and that we're taking $8$ bulbs directly from the manufacturing process, in which case method I is again correct. $\endgroup$ – Cameron Buie Oct 27 '13 at 13:59
  • $\begingroup$ Still, the suggestion is that the selection is made without replacement, and the phrasing indicates a single lot. What is it that leads you to believe that the answer from method I is correct? $\endgroup$ – Cameron Buie Oct 27 '13 at 14:03
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The problem says: "A LOT of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non defective bulbs". It doesn't say that there is only 100 bulbs (10 defective and 90 non defective), it means that in every group of 100 bulbs should be 10 defective and 90 non defective bulbs by probability, in other words - the rate of defective bulbs is 10/100 and the rate of non defective bulbs is 90/100. So consider that you have an infinite number of bulbs which consists of 10% of defective and 90% of non defective bulbs. That's why the 1st solution gives you the right answer :) The problem could also say "A lot of 10 bulbs from a manufacturing process is known to contain 1 defective and 9 non defective bulbs", it doesn't matter on the numbers, but on the rate of bulb types.

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  • $\begingroup$ I must disagree. The problem describes a lot, as in a particular set of bulbs. It says nothing about all such lots (though of course, the larger the lot, the closer the lot's defect rate should be to the overall defect rate). Had the problem said "A manufacturing process is known to produce defective bulbs $10\%$ of the time," then method I would be spot on. $\endgroup$ – Cameron Buie Oct 26 '13 at 16:24
  • $\begingroup$ well, then the question is not formulated correctly, because the first solution returns the right answer :) $\endgroup$ – Gintas K Oct 27 '13 at 13:44
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    $\begingroup$ In hindsight, it is not made clear whether the bulbs are being selected with or without replacement, so as phrased, it's a bad question. $\endgroup$ – Cameron Buie Oct 27 '13 at 14:00
  • $\begingroup$ The word "lot" in the question is not used in its colloquial informal meaning of "many", but in the sense of "batch" (dictionary also has "an item or set of items for sale at an auction"). $\endgroup$ – ShreevatsaR Apr 2 '14 at 18:34
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Method II is the correct method. However you are drawing $8$ bulbs, not $9$, and so the answer becomes $$1 - \left(\frac{90}{100} \cdot \frac{89}{99} \cdot \ldots \cdot \frac{83}{93}\right)$$

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  • $\begingroup$ thats okay.. that was a typing mistake from my side.. but this is not the answer ...if you see the answer in method I... $\endgroup$ – sachin Oct 24 '13 at 14:49
  • $\begingroup$ Your method I does not give the correct answer. There you are assuming that you draw with replacement, i.e. the chances of drawing a defective bulb stay the same for all $8$ bulbs that you draw. This is not the case though. $\endgroup$ – Arthur Oct 24 '13 at 14:52
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This is a classic case of how conditional probability changes the independence of events, i.e. draws from the lot.

Method 2 assumes that you look at each bulb when you draw it, which means that you gain information about the remaining bulbs, and the next draw is now conditionally dependent on the previous draws (e.g. drawing 7 bad bulbs in a row changes the odds of drawing a bad bulb from 10/100 = 10% to 3/93 = 3.23%.

The problem is asking for 8 bulbs selected at random. This means that you either scooped out 8 bulbs at once or drew 8 bulbs individually with a blindfold on, and then took the blindfold off and looked at all 8 together. This keeps the draws independent. This is what the problem is asking. Only Method 1 is correct; Method 2 answers a different question.

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protected by hardmath Dec 14 '16 at 23:17

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