Question

232 CHAPTER9Introducing Probability (b) What is the probability that a randomly chosen postmenopausal woman has some problem with low bone density (osteopenia or osteoporosis)? 9.35 Hepatitis C. The CDC provides the breakdown of the sources of infection leading to hepatitis C in Americans. Here are the probabilities of each infection source for a randomly chosen individual with hepatitis C Probability Source of infection 0.60 0.15 Intravenous drug use Unprotected sex Transfusion (before screening) Unknown or other Occupational 0.10 0.11 (a) What is the probability that a person with hepatitis C was infected in the course of his or her professional occupation? What is the probability that a person with hepatitis C was infected through a known risky behavior (intravenous drugs or unprotected sex)? (b) 9.36 Birth order. The National Vital Statistics database re- ports all births in 2011 by the reported birth order of the child (first child of the mother, second, etc.). Here is how the 3,923,381 documented births break down: Birth order First child Second child Third child Percent 40.2 31.6 16.5

Answer 9.35

Answer 1

Since,we know that the sum of probabilities always lies between 0 and 1.

So, for here we are given probabilities of four sources of infection and the last one i,e, "occupational" is unknown.

Therefore, we'll follow law of probability that the sum of total probability should be equals to 1.

(a) $\implies$ Let the Probability of "occupational" be 'X'

Therefore, P(Intravenous drug) + P(Unprotected sex) + P(Transfusion) + P(unknown or other) + P(occupational) = 1

$\implies$0.60 + 0.15 + 0.10 + 0.11 + X = 1

$\implies$0.96 + X = 1

$\implies$X = 1 - 0.96

$\implies$X = 0.04

that means, P(person getting Hepatitis C due to professional occupation) = 0.04

(b) When the probability that one of the two events occur then, there probabilities get added.

For an instance, Let A ,B be two events and with probabilities P(A) and P(B) respectively

then for P(getting A or B) = P(A) + P(B)

Similarly, P(person getting Hepatitis c due to intravenous drug or unprotected sex) = P(intravenous drug) + P(unprotected sex)

$\implies$ P(person getting Hepatitis c due to intravenous drug or unprotected sex) = (0.60) + (0.15) = 0.75