A lot of 100 semiconductor chips contains 20 that are defective. (a) two are selected, at random, without replacement, from the lot. determine the probability that the second chip selected is defective. (b) three are selected, at random, without replacement, from the lot. determine the probability that all are defective.
Given: all selections are without replacement Let D=event that chip is defective N=event that chip is normal note that D and N are complementary. x=undefined event (a) Second is defective Probability for second chip to be defective P(xD)=P(ND)+P(DD) P(ND)=(80/100)*(20/99)=1600/9900 P(DD)=(20/100)*(19/99)=380/9900 => P(xD)=(1600+380)/9900=1980/9900=1/5
(b) All three are defective P(DDD)=(20/100)*(19/99)*(18/98)=19/2695
Note: in probabilities, it is preferable to use the exact values (i.e. fractions).
