Exercise 13.3 · Q5: Blood test: has the disease, given a positive result, $P = \frac{22}{133}$

Basics you need

Solution

1. Name the events

   Let

   • $E_1$⁠: the person tested has the disease;
   • $E_2$⁠: the person tested does not have the disease (is healthy);
   • $A$⁠: the test result is positive.

   $E_1$ and $E_2$ form a partition of the sample space: $E_2$ is the complement of $E_1$⁠, and both have nonzero probability.

   "Has the disease" and "healthy" are the two possible causes; the positive result is what we observe. An event and its complement form a partition when both have nonzero probability (Section 13.5.1), as here: $\frac{1}{1000}$ and $\frac{999}{1000}$⁠.
   $\{E, E'\}$ is a partition when $E$ is neither impossible nor certain. Lesson: Bayes' theorem

2. Every number as a fraction

   $$\begin{aligned} P(E_1) &= 0.1\% = \frac{0.1}{100} = \frac{1}{1000}, \\ P(E_2) &= 1 - \frac{1}{1000} = \frac{999}{1000}, \\ P(A|E_1) &= 99\% = \frac{99}{100}, \\ P(A|E_2) &= 0.5\% = 0.005 = \frac{1}{200}. \end{aligned}$$
   "0.1 percent of the population actually has the disease" gives the chance that a person picked from the population has it. "99% effective in detecting" is about people who have the disease, so it is $P(A|E_1)$⁠. The false-positive rate is about healthy people, so it is $P(A|E_2)$⁠; the book itself writes it as the probability 0.005. Fractions keep everything exact: $0.005 = \frac{5}{1000} = \frac{1}{200}$⁠.
   $x\% = \frac{x}{100}$⁠, and $P(E') = 1 - P(E)$⁠. Basics: rules of probability

3. The chance of a positive result

   Since $E_1$ and $E_2$ form a partition, by the theorem of total probability

   $$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac{1}{1000}\cdot\frac{99}{100} + \frac{999}{1000}\cdot\frac{1}{200} \\ &= \frac{99}{100000} + \frac{999}{200000} \\ &= \frac{198}{200000} + \frac{999}{200000} \\ &= \frac{1197}{200000}. \end{aligned}$$
   A positive result happens through a true positive or a false positive, never both, so the two path probabilities add. To add them, write $\frac{99}{100000}$ as $\frac{198}{200000}$⁠.
   Multiply along a path, add the paths. Basics: tree diagrams

4. Has the disease, given positive: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(E_1|A) &= \frac{P(E_1)P(A|E_1)}{P(A)} \\ &= \frac{198}{200000} \div \frac{1197}{200000} \\ &= \frac{198}{1197}. \end{aligned}$$

   Both 198 and 1197 are divisible by 9: $198 = 9 \times 22$ and $1197 = 9 \times 133$⁠. So

   $$P(E_1|A) = \frac{22}{133} \approx 0.165.$$
   The top is the true-positive path; the bottom is all positive results. The common denominator 200000 cancels. No further cancelling is possible: $22 = 2 \times 11$ and $133 = 7 \times 19$ have no common factor.
   Simplify by the highest common factor, here 9. Basics: working with fractions

5. What the answer means

   Only about 16.5% of people who test positive actually have the disease. The other positives (about 83.5%) are false alarms: $P(E_2|A) = \frac{999}{1197} = \frac{111}{133}$⁠, and $\frac{22}{133} + \frac{111}{133} = 1$⁠.

   The test is not useless: it raised the chance of disease from $\frac{1}{1000}$ to $\frac{22}{133}$⁠, more than 160 times. But the disease is so rare that a positive result calls for a second test, not a diagnosis.

   A healthy person is about 1000 times as common as a sick one, and a false positive happens for 1 healthy person in 200. Those false positives outnumber the true ones about 5 to 1.
   $P(E_1|A) + P(E_2|A) = 1$ for a partition. Lesson: Bayes' theorem

Every fact used, and where it comes from