Which Probability Calculation is Correct for Defective Bulbs in a Sample?

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Homework Help Overview

The discussion revolves around calculating the probability of selecting non-defective bulbs from a box containing 100 bulbs, of which 10 are defective. Participants are exploring different approaches to determine the probability that none of the 5 sampled bulbs is defective.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants present two methods for calculating the probability: one using combinations without replacement and another using a probability approach assuming replacement. Others question the independence of events when selecting bulbs and whether the sampling method affects the calculations.

Discussion Status

The discussion is ongoing, with participants examining the implications of sampling with and without replacement. Some have provided insights into the differences between the two methods, while others express confusion regarding the correct interpretation of the problem and the provided solutions.

Contextual Notes

There is a noted discrepancy between the answers derived from different methods and the solutions found in external sources, leading to further questioning of the assumptions made in the problem setup.

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Homework Statement



In a box containing 100 bulbs, 10 are defective. The probability that out of a
sample of 5 bulbs, none is defective is?

The Attempt at a Solution



I see two approaches to this problem.

1) Out of 90 non-defective bulbs, we can chose 5 in 90C5 ways.
There are a total no. of 100C5 ways.

So required probability =90C5/100C5=0.5838

2) Out of the sample of 5 bulbs,
Probability of a non-defective bulb = 90/100=9/10
In 5 bulbs, required probability = (9/10)5=0.59049

The two answers don't differ by much. Which one is correct and why?
 
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Abdul Quadeer said:
2) Out of the sample of 5 bulbs,
Probability of a non-defective bulb = 90/100=9/10
In 5 bulbs, required probability = (9/10)5=0.59049

You can multiply the probabilities of independent events to calculate the probability of all of them occurring together. Is pulling a second non-defective light bulb independent of pulling a first one?

In other words, suppose you are picking the second lightbulb from the box. Does the probability depend on the condition of the first lightbulb?
 
In the selected sample, you don't take away the bulbs after picking them. In other words, the bulbs are replaced in the sample. So they are independent events.
 
So when they say "a sample of 5 bulbs", you can have the same bulb multiple times in the sample?
 
Abdul Quadeer said:

Homework Statement



In a box containing 100 bulbs, 10 are defective. The probability that out of a
sample of 5 bulbs, none is defective is?

The Attempt at a Solution



I see two approaches to this problem.

1) Out of 90 non-defective bulbs, we can chose 5 in 90C5 ways.
There are a total no. of 100C5 ways.

So required probability =90C5/100C5=0.5838

2) Out of the sample of 5 bulbs,
Probability of a non-defective bulb = 90/100=9/10
In 5 bulbs, required probability = (9/10)5=0.59049

The two answers don't differ by much. Which one is correct and why?
This is selection without replacement. In (2) you are calculating the probability with replacement. That is, as if you take a bulb, test it, put it back in the box and choose again, with a (slight) chance of getting the same bulb again.

Instead you could argue that, at first, there are 90 non-defective bulbs out of 100 so the chance that the first bulb selected is non-defective is 90/100= .9. But then there are 89 non-defective bulbs left among 99 bulbs. The chance of selecting a non-defective bulb the second time is 89/99, not 90/100 again.

The probability of selecting 5 non-defective bulbs is (90/100)(89/99)(88/98)(87/97)(85/96)= 0.5838 as in (1).
 
HallsofIvy said:
This is selection without replacement. In (2) you are calculating the probability with replacement. That is, as if you take a bulb, test it, put it back in the box and choose again, with a (slight) chance of getting the same bulb again.

Instead you could argue that, at first, there are 90 non-defective bulbs out of 100 so the chance that the first bulb selected is non-defective is 90/100= .9. But then there are 89 non-defective bulbs left among 99 bulbs. The chance of selecting a non-defective bulb the second time is 89/99, not 90/100 again.

The probability of selecting 5 non-defective bulbs is (90/100)(89/99)(88/98)(87/97)(85/96)= 0.5838 as in (1).

I understood your explanation.
But the thing is that, in my text the answer given is (0.9)^5. Not only this, I searched this question in google books and found that everywhere the answer is same as above.

ex:

18. The probability of getting a defective bulb from the box, is 1 / 10.

Hence using binomial distribution,

the required probability , is (0.9)^5.

Hence the option is C.

Source - http://creatorstouchglobal.com/gm/index.php?option=com_content&view=article&id=63&Itemid=121
 
Last edited by a moderator:
It is really seems wrong on that website. If the question was "What is the probability of pulling a non-defective bulb 5 times if the bulbs are replaced after each drawing" then the answer would be (0.9)^5, but "a sample of 5 bulbs" strongly implies no replacement.
 
hmmm...its possible that the answer is wrong.
 

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