Question

Paragraph 2-114. A lot of 100 semiconductor chips contains 10 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 selêcted, at random, without replacement, from the lot. Determine the probability that all are defective.

Answer 1

There are total 100 chips out of that 10 are defective and the remaining 100 - 10 = 90 are non defective.

(a) 2 chips are randomly selected without replacement.

To find the probability of second selected defective, there are total 99 chips. Since one chip is already selected at first and kept aside. So there are total 99 chips after selecting the first.

And there are total 10 defective chips.

Therefore

Therefore, the probability that the second chip selected defective is 0.10101

(b) 3 chips are selected randomly without replacement.

P(All are defective) -P(First de fective) P(Second defective) * P(Third defective)

P(Second def ective) Total number of Number of de fective after selecting first Total number of chips after selecting first 9

Therefore, the probability that all are defective is 0.00074