Alot 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?

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

If you need an equation, it would be y=2x+8, y would be how long it grew, x being weeks.if you need how long he was asleep, it would be 4 weeks if he started with no beard, the equation would be x=8÷2if he did have a beard before hand, the equation would be 2w+x=8