A test for a disease is 95% accurate, meaning it correctly identifies both positive and negative cases 95% of the time. The disease is present in 1% of the population. If a person tests positive, what is the probability that they actually have the disease?

$0.16$

The correct answer is Option (3) → $0.16$ ##

Using Bayes' theorem:

$P(\text{Disease} \mid \text{Positive}) = \frac{P(\text{Positive} \mid \text{Disease}) \times P(\text{Disease})}{P(\text{Positive})}$

Where:

• $P(\text{Positive} \mid \text{Disease}) = 0.95$
• $P(\text{Disease}) = 0.01$
• $P(\text{Positive} \mid \text{No Disease}) = 0.05$

Now, calculate $P(\text{Positive})$:

$P(\text{Positive}) = P(\text{Positive} \mid \text{Disease}) \cdot P(\text{Disease}) + P(\text{Positive} \mid \text{No Disease}) \cdot P(\text{No Disease})$

$= (0.95 \times 0.01) + (0.05 \times 0.99) = 0.059$

Thus, $P(\text{Disease} \mid \text{Positive}) = \frac{0.95 \times 0.01}{0.059} \approx 0.16$