a manufacture of halogen bulbs knows that 3% of the production of their 100W bulbs will dective

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Ryan S. | Mathematics and StatisticsMathematics and Statistics

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.

(12*11*10*9*8*7*6*5*4*3*2*1)/(5*4*3*2*1*7*6*5*4*3*2*1)=792

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

792*.03^5*.97^7=0.00001555

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.

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