Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards, and 20 are selected without replacement for functional testing. a) If 20 cards are defective, what is the probability that at least 1 defective card is in the sample? b) If 5 cards are defective, what is the probability that at least 1 defective card appears in the sample?

a) 0.9644 or 96.44%

b) 0.5429 or 54.29%

Step-by-step explanation:

a) The probability that at least 1 defective card is in the sample P(A) = 1 - probability that no defective card is in the sample P(N)

P(A) = 1 - P(N) 1

Given;

Total number of cards = 140

Number selected = 20

Total number of defective cards = 20

Total number of non defective cards = 140-20 = 120

P(N) = Number of possible selections of 20 non defective cards ÷ Number of possible selections of 20 cards from all the cards.

P(N) = 120C20/140C20 = 0.0356

From equation 1

P(A) = 1 - 0.0356

P(A) = 0.9644 or 96.44%

b) Using the same method as a) above

P(A) = 1 - P(N) 1

Given;

Total number of cards = 140

Number selected = 20

Total number of defective cards = 5

Total number of non defective cards = 140-5 = 135

P(N) = 135C20/140C20 = 0.457

From equation 1

P(A) = 1 - 0.4571

P(A) = 0.5429 or 54.29%

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