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.

$$\begin{gathered} \therefore P\left(E_1\right)=0.1\%=0.001 \\ \mathrm{Now}, \\ P\left(A/E_1\right)=99\%=0.99 \\ P\left(A/E_2\right)=0.5\%=0.005 \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005} \\ =\frac{990}{5985}=\frac{22}{133} \end{gathered}$$