Hi,



A production line manufactures light bulbs with a defective rate of 0.05. What is P(in a batch of 20 light bulbs, 2 or more of them are defective)?



Note: The probability of x successes in n trials is: 

P = nCx*  where p and q are the probabilities of success and failure respectively. 

In this case p = .05  and q = .95

nCx = 

P(2 or more defective) = 1 - P(none) - P(one) 

                       = 1 -(.95)^20 - 20(.05)^1*(.95)^19

                       = 1 - .3585 - .3774 = .2641

P( 1 or 2 are defective) = .3774 + .1887 = .5661 

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