Question

An unhealthy triangular-shaped region in an old industrial city once had a lot of chemical industry. Two percent of the children in the city live in the triangle. Fourteen percent of these test positive for excessive presence of toxic metals in the tissues. The rate of positive tests for children in the city, not living in the triangle, is only 1%.
(a) What is the probability that a child who lives in the city, and who is chosen at random, both lives in the Triangle and tests positive?
(b) What is the probability that a child living in the city, chosen at random, tests positive?

Answer 1

You may or may not care what the answer is at this point and/or already know how to solve these problems by now, but in any case here's how you solve it:

The probability of separate events occurring at the same time is given by multiplying those two probabilities together. In this problem, there are two separate events which occur:
1. living in or out of the triangle
2. testing positive or negative

a.) P of living in the Triangle and testing positive = (Probability of living in the Triangle) * (Probability of testing positive)

P = (2%) * (14%)
P = (0.02) * (0.14)
P = 0.0028 or 0.028% chance of living in the Triangle and testing positive

b.) P of testing positive = P of testing positive in Triangle + P of testing positive in the rest of the city

P = (2%) * (14%) + (98%) * (1%)
P = (0.02) * (0.14) + (0.98) * (0.01)
P = 0.0028 + 0.0098
P = 0.0126 or 1.26% chance of testing positive in the city

Answer 2

*correction. on part a.) I meant 0.28%