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%
i believe the answer is d.
hope it you! : d