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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?

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You say $0.2$, but the answer in the back of your book is right if it's $0.02$. See below. –  Michael Hardy Nov 12 '12 at 17:15

2 Answers 2

up vote 4 down vote accepted

$$ \begin{array}{cccccccccccc} & & & & & \bullet \\[15pt] & & & & \swarrow0.2 & & \searrow0.8 \\[15pt] & & & D & & & & \overline{D} \\[15pt] & & & & & & \swarrow & & \searrow \\[15pt] & & & & & & \text{positive} & & \text{negative} \\[15pt] & & & & & \swarrow & & \searrow \\[15pt] & & & & D & & & & \overline{D} \end{array} $$ 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.

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That makes a lot of sense to me, the only thing is that they are giving an answer of $16.90 at the end of the book... Do you have any idea how they are coming to that answer? –  TopGunCpp Nov 12 '12 at 16:04
    
+1 for the plot :) –  mythealias Nov 12 '12 at 16:05
    
I deleted this answer, then edited further, then un-deleted it. I initially omitted the fact that everyone who is ill also takes the second test. –  Michael Hardy Nov 12 '12 at 16:50
    
If $\$16.90$ is given as the answer, it's got to be a weighted average of $\$110$ and $\$10$ for some weights, say $w$ and $1-w$. So $110w+10(1-w)=16.90$. That says $100w + 10=16.90$, so $100w=6.9$, and $w=0.069$. That would mean only $6.9\%$ of the population take the second test. To me that makes no sense, since all of the $20\%$ who are ill would take the second test. –  Michael Hardy Nov 12 '12 at 16:55
    
OK, here's the bottom line: if $P(D)=0.02$, i.e. $2\%$ rather than $20\%$ have the disease, then the answer is $\$16.90$. –  Michael Hardy Nov 12 '12 at 17:14

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.

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