Question

A lot of 101 semiconductor chips contains 25 that are defective.

(a)	Two are selected, one at a time and without replacement from the lot. Determine the probability that the second one is defective.

(b)	Three are selected, one at a time and without replacement. Find the probability that the first one is defective and the third one is not defective.

Answer 1

(a)

Number of non-defective semiconductor chips = 101 - 25 = 76

Probability that the second one is defective = P(First one is defective) * P(Second one is defective) + P(First one is not defective) * P(Second one is defective)

= (25/101) * (24/100) + (76/101) * (25/100)

= 0.2475248

(b)

Probability that the first one is defective and the third one is not defective = P(First one is defective) * P(Second one is defective) * P(Third one is not defective) + P(First one is defective) * P(Second one is not defective) * P(Third one is not defective)

= (25/101) * (24/100) * (76/99) + (25/101) * (76/100) * (75/99)

= 0.1881188