<PRE><B>Question 417737</B>
based on the recent data it is found that 5% of the out put has one or more defects . in spot checking some parts , inspector randomly selects two parts find the probabilities that
Sample space: D means defective; N means not defective
DD,DN,ND,NN 
a. the first part is defect free::0.95*0.05 + 0.95*0.95 = 0.95
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b. the second part is defect free::same as above: 0.95
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c. both parts are defect free: 0.95^2 = 0.9025
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d. one of the parts is acceptable::2(0.05*0.95) = 0.095
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e. at least one part is acceptable:: 1-(0.05)^2 = 0.9975
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Cheers,
Stan H. 
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