I'm assuming the limit is supposed to be
Multiply the numerator by its conjugate, and do the same with the denominator:
so that in the limit, we have
Factorize the first term in the denominator as
The 
terms cancel, leaving you with
and the limand is continuous at 
, so we can substitute it to find the limit has a value of -1/18.
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
Answer:
option D 12 is correct..
in second question the student didn't multiply -3 with 4. that's the mistake he did.
hope it helps
