Question

A lot of 100 semiconductor chips contains 15 that are defective. three chips are selected at random from the lot without replacement. what is the probability that the third chip is defective?

Answer 1

D=event that chip selected is defective
d=event that chip selected is NOT defective

Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154

Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20