Statistics Homework Question I tried to got my my TA's office hours today but someone hogged him for 45 minutes and I didn't get a chance to ask my question. Hopefully someone can shed some light on this for me.
Note that this is a question about diagnosing people with a disease
$P(D)$ is the probability the person has the disease.
Assume that for a randomly selected person, $P(D) = 0.2$, $P(\text{Positive test}\mid D) = 1$. $P(\text{Positive test}\mid\overline{D}) = 0.05$, so that the inexpensive test only gives false positive, and not false negative results.
Suppose that this inexpensive test costs $\$10$. If a person tests positive then they are also given a more expensive test costing $\$100$, which correctly identifies all persons with the disease. What is the expected cost per person if a population is tested for the disease using the inexpensive test followed, if necessary, by the expensive test?
 A: $$
\begin{array}{cccccccccccc}
& & & & & \bullet \\[15pt]
& & & & \swarrow0.2 & & \searrow0.8 \\[15pt]
& & & D & & & & \overline{D} \\[15pt]
& & & & & & \swarrow & & \searrow \\[15pt]
& & & & & & \text{positive} & & \text{negative} \\[15pt]
& & & & & \swarrow & & \searrow \\[15pt]
& & & & D & & & & \overline{D}
\end{array}
$$
Follow this tree.  Everybody pays $\$10$ for the first test.
Then $5\%$ of the aformentioned $80\%$ of the population pays $\$100$ for the second test.  
The $20\%$ who are ill also pay the $\$100$ for the second test.
So
$$
\$10 + (0.2)\$100 + (0.05)(0.8)\$100 = \$10 + \$4 = \$34.
$$
Another p.o.v.: You have $100$ people.  Of those, $20$ are ill.  They take both tests, paying $\$110$ each.  Of the other $80$ people, $5\%$ take both tests, paying $\$110$.  That's $4$ people.  They other $76$ test negative on the cheap test, and pay only $\$10$ each.  So
$$
24\cdot\$110 + 76\cdot\$10 = \$3400.
$$
So the average is $\$34$.
Later note: You say the answer at the end of the book is $\$16.90$.  That would be correct if $P(D)=0.02$ rather than $0.2$.  Just go through the above with those numbers instead of what we used.
A: Draw yourself a probability tree. On the first fork we have $P(D) = 0.2$ and $P(D') = 0.8$. On the second set of forks we have $P(+|D) = 1$, $P(-|D) = 0$, $P(+|D')=0.05$ and $P(-|D') = 0.95$. It follows that $P(+ \wedge D) = 0.2 \times 1 = 0.2$, $P(- \wedge D) = 0.2 \times 0 = 0$, $P(+ \wedge D') = 0.8 \times 0.05 = 0.04$, and $P(- \wedge D') = 0.8 \times 0.95 = 0.76$.
The expected value can be found by multiplying the probability by the cost. I get
$$E[X] = 0.2 \times 110 + 0 \times 10 + 0.04 \times 110 + 0.76 \times 10 = 34 \, .$$
Thus, you can expect to pay 34 dollars per person.
