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