Question

A laboratory blood test is 99% effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer 1

Let E₁ and E₂ denote the events that a person has a disease and a person has no disease, respectively.

E₁ and E₂ are complimentary to each other.

∴ P (E₁) + P (E₂) = 1

⇒ P (E₂) = 1 − P (E₁) = 1 − 0.001 = 0.999

Let A denote the event that the blood test result is positive.

\[\therefore P\left( E_1 \right) = 0 . 1 % = 0 . 001 \]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = 99 % = 0 . 99\]
\[P\left( A/ E_2 \right) = 0 . 5 % = 0 . 005\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{0 . 001 \times 0 . 99}{0 . 001 \times 0 . 99 + 0 . 999 \times 0 . 005}\]
\[ = \frac{990}{5985} = \frac{22}{133}\]