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What is the probability that exactly 5 bulbs in a carton of 12 bulbs will be defective?

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This is an example of a binomial distribution. Each of the 12 bulbs in the carton is a "trial" with a 3% probability of being defective. We need to know how many ways we can have 5 bulbs be defective among 12. This is given by the combination function which can be written as n!/(k!(n-k)!) where n is the number of elements and k is the size of the combination. In this case n=12 and k=5. 
So we have 792 combinations possible where there are 5 defective bulbs and 7 good bulbs. The probability for one of these is .03^5*.97^7. So the total probability is
By dividing 5 by .03, you find that it would take at least 166 bulbs to have 5 defective bulbs.  Using a proportion to bring (5 out of 166 bulbs) down to (x out of twelve bulbs), you may find a probability of 0.36.
Please respond if this is not helpful.  This answer was given to ignite some logical thought.


I hope that you will learn about the binomial distribution.
Michael F.