conditional probability
posted by Tifini on .
Among a group of people, 10% are from NYC and 90% are not. All people from NYC are under 18 years old, and 60% of those not from NYC are also under 18 years old. One person is chosen at random, and he or she is under 18. What is the probability that the person chosen is from NYC? Round your answer to two decimal places.
how would i set this problem up and solve it??

Define events
N=from NYC, hence N'=not from NYC
A=adolescents under 18, hence A'=over 18
Given:
P(AN)=1.0 (prob. of under 18 given person is from NYC)
P(AN')=0.6 (prob. of under 18 given given person is from elsewhere)
First calculate
probability of finding a person under 18,
P(A)=P(AN)*P(N)+P(AN')*P(N') [ Bayes thm]
=1.0*0.1+0.6*0.9
=0.64
Now we look for the probability that someone is from NYC given she's under 18
P(NA)
=P(N∩A)/P(A) [cond. prob.]
=P(A∩N)/P(A) [commutativity]
=P(AN)P(N)/P(A) [Bayes Thm]
=1.0*0.1/0.64
=0.156