A disease is known to be present in 2% of a population - Edexcel - A-Level Maths: Statistics - Question 1 - 2008 - Paper 2

To find the total probability that the test is positive, we can use the law of total probability:

$P(Positive \, Test) = P(Disease) \cdot P(Positive \, Test | Disease) + P(No \, Disease) \cdot P(Positive \, Test | No \, Disease)$

Substituting the known probabilities:

$P(Positive \, Test) = 0.02 \cdot 0.95 + 0.98 \cdot 0.03$

Calculating this gives:

$P(Positive \, Test) = 0.019 + 0.0294 = 0.0484$

To find the probability that a person does not have the disease given that the test is positive, we use Bayes' theorem:

$P(No \, Disease | Positive \, Test) = \frac{P(Positive \, Test | No \, Disease) \cdot P(No \, Disease)}{P(Positive \, Test)}$

Using the previously calculated values:

$P(No \, Disease | Positive \, Test) = \frac{0.03 \cdot 0.98}{0.0484}$

Calculating this gives:

$\approx 0.607 \text{ (or about 60.7 ext{\%}})$

The test is not very useful since, even if the test result is positive, there is still a high probability (about 60.7%) that the individual does not have the disease. This low specificity indicates that the test may lead to a significant number of false positives, which could cause unnecessary anxiety and possibly lead to further invasive testing.