Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards. A sample of 20 cards are selected from the lot without replacement for functional testing. (a) If 20 cards are defective, what is the probability that at least one defective card appears in the sample

The probability that at least one defective card appears in the sample

P(D) = 0.9644 or 96.44%

Step-by-step explanation:

Given;

Total number of cards t = 140

Number of defective cards = 20

Number of non defective cards x = 140-20 = 120

The probability that at least one defective card = 1 - The probability that none none is defective

P(D) = 1 - P(N) 1

For 20 selections; r = 20

-- 20 cards are selected from the lot without replacement for functional testing

The probability that none none is defective is;

P(N) = (xPr)/(tPr)

P(N) = (120P20)/(140P20)

P(N) = (120!/(120-20)!)/(140!/(140-20)!)

P(N) = (120!/100!)/(140!/120!) = 0.035618370821

P(N) = 0.0356

The probability that at least one defective card appears in the sample is;

P(D) = 1 - P(N) = 1 - 0.0356 = 0.9644

P(D) = 0.9644 or 96.44%

Note: xPr = x permutation r