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title: '17'

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To find the probability that a person has the disease given that the test result is positive, use Bayes' theorem.

Let's define the following terms:

$A:$ The event that a person has the disease.
$B:$ The event that the test result is positive.

To find $P(A | B)$, which is the probability that a person has the disease given a positive test result.

According to Bayes' theorem[1]:
$$P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}$$

Where,
$P(B | A)$: The probability that the test is positive given that a person has the disease is 95%, so $P(B | A) = 0.95$.

$P(A)$: The probability that a person has the disease is $0.5\%$, so $P(A) = 0.005$ ($0.5\%$ as a decimal).

To calculate $P(B)$, use the law of total probability [2]:

$P(B) = P(B | A) \cdot P(A) + P(B | \text{not } A) \cdot P(\text{not } A)$

Where:
$P(B | \text{not } A)$: The probability that the test is positive given that a person doesn't have the disease is the false positive rate, which is $1\%$, so $P(B | \text{not } A) = 0.01$.

$P(\text{not } A)$: The probability that a person doesn't have the disease is $100\% - 0.5\% = 99.5\%$, so $P(\text{not } A) = 0.995$.

Now, calculate $P(B)$:

$P(B) = (0.95 \cdot 0.005) + (0.01 \cdot 0.995) = 0.00475 + 0.00995 = 0.0147$

Now, you can use Bayes' theorem to find $P(A | B)$:
$$P(A | B) = \frac{0.95 \cdot 0.005}{0.0147} \approx 0.3231$$

So, the probability that a person has the disease given a positive test result is approximately $0.3231$ or $32.31\%$.