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 func- tional 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? (montgomery 97) montgomery, douglas c. applied statistics and probability for engineers, 6th edition. wiley, 2013-10-21. vitalbook file. the citation provided is a guideline. please check each citation for accuracy before use.
A) If 20 cards are defective, 120 are not. There are C(120, 20) ≈ 2.946·10²² ways to choose 20 cards of which none are defective. There are C(140, 20) ≈ 8.272·10²³ ways to choose 20 cards. So, the probability of choosing 20 cards of which none are defective is about 2.946/82.72 ≈ 0.03562. This means the probability that at least one is defective is 1 - 0.03562 ≈ 0.96438
Step-by-step explanation: All you'd have to do in order to achieve this answer is by dividing 410 by 40 which gives you the answer of 10.25. In order to fit everyone they'll need a rounded number of buses; equalling 11.
