Suppose that strep throat affects 2% of the population and a test to detect it produces an accurate result 99% of the time. a. Create a two-way table for 10,000 people who were tested. b. What is the probability that someone who tests positive actually has strep throat?

(a) We consider 10,000 people who are being tested. 2% of the population has strep throat and thus $2\mathrm{\%}\times 10,000=0.02\times 10,000=200$ people have strep throat (while the remaining $10,000-200=9,800$ people have no strep throat). 99% of the population has an accurate test result and thus 99% of the people with strep throat test positive ${\displaystyle (99\mathrm{\%}\times 200=0.99\times 200=198\text{people}),}$ while 99% of the people with no strep throat test negative $(99\mathrm{\%}\times 9800=0.99\times 9800=9702\text{people}).$ Since 198 of the 200 people with strep throat test positive, thus $200-198=2$ of the 200 people with strep throat test negative. Since 9702 of the 9800 people with no strep throat test negative, thus $9800-9702=98$ of the 9800 people with no strep throat test positive. $\begin{array}{cccc}& \text{Have strep}& \text{Do not have strep}& Total\\ TestPositive& 198& 98& 296\\ TestNegative& 2& 9702& 9704\\ Total& 200& 9800& 10,000\end{array}$

(b) We note that 296 people test positive, because 296 is mentioned in the row ” Test Positive” and in the column ”Total” of the table. We also note that 198 people test positive and have strep, because 198 is mentioned in the row ” Test Positive” and in the column ”Have strep” of the table. Thus 198 of the 296 people who test positive have strep throat. The probability is the number of favorable outcomes divided by the number of possible outcomes: $P(HaveStrep|TestPositive)=\frac{\mathrm{\#}offavorableoutcomes}{\mathrm{\#}ofpossibleoutcomes}$
${\displaystyle =\frac{198}{296}}$
${\displaystyle =\frac{99}{148}}$
${\displaystyle \approx 0.6689}$
${\displaystyle =66.89\mathrm{\%}}$