A diagnostic test for a certain disease has a sensitivity of 95% and a false positive rate of 10%. If the prevalence of the disease in the population is 20%, what is the probability that a person who

Let $D$ be the event that a person has the disease, and $+$ be the event that the test is positive.

• Sensitivity (probability of testing positive given the disease): $P(+|D) = 0.95$
• False positive rate (probability of testing positive given no disease): $P(+|\neg D) = 0.10$
• Disease prevalence (prior probability of having the disease): $P(D) = 0.20$

The probability of not having the disease, $P(\neg D)$, is $1 - P(D)$.

$$P(\neg D) = 1 - P(D) = 1 - 0.20 = 0.80$$

We want to find the probability of having the disease given a positive test, $P(D|+)$. Bayes' theorem states:

$$P(D|+) = \frac{P(+|D) \cdot P(D)}{P(+|D) \cdot P(D) + P(+|\neg D) \cdot P(\neg D)}$$

Substitute the known probabilities into the formula:

$$P(D|+) = \frac{0.95 \times 0.20}{(0.95 \times 0.20) + (0.10 \times 0.80)}$$

Calculate the terms:

$$P(D|+) = \frac{0.19}{0.19 + 0.08}$$ $$P(D|+) = \frac{0.19}{0.27}$$ $$P(D|+) \approx 0.7037$$

Rounding 0.7037 to two decimal places gives 0.70.

The probability of having the disease given a positive test is approximately 0.70.

The correct option is d) 0.70.