Question

A laboratory blood test is \[99\% \] effective in detecting certain diseases when it 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

Hint: According to the question, there are mainly two kinds of events which are when a person has the disease and a person does not have any disease. When we take these two events as complementary events, and by taking the application of Bayes theorem, we can get the answer.

Complete step-by-step answer:
Let us consider those two events as a person having a disease to be \[E1\], and a person having no disease to be \[E2\].
Now,\[E1\] and \[E2\] both are complementary events. So, we get:
\[ \Rightarrow P(E1) + P(E2) = 1\]
\[ \Rightarrow P(E2) = 1 - P(E1) = 1 - 0.001 = 0.999\]
Therefore, the probability of a person having any disease is:
\[P(E1) = 0.1\% \; = 1000.1 = 0.001\]
Now, let us consider that the blood test result is positive for an event called A.
From the question, the probability of the person having any disease, if test results is positive will be:
\[ = 99\% = 99/100 = 0.99 = P(A\mid E1)\]
Similarly, the probability of a person having no disease, but the result is positive will be:
\[ = 0.5\% = 0.5/100 = 0.005 = P\left( {A\mid E2} \right)\]
The probability of a person having disease if the test result is positive is
\[P\left( {E1\mid A} \right)\]
So, by applying Bayes theorem, we get:
\[P(E1\mid A) = \dfrac{{P(E1) \times P(A\mid E1)}}{{P(E1) \times P(A\mid E1) + P(E2) \times P(A\mid E2)}}\]
\[ = \dfrac{{0.001 \times 0.99}}{{0.001 \times 0.99 + 0.999 \times 0.005}}\]
\[ = \dfrac{{0.00099}}{{0.00099 \times 0.004995}}\]
\[ = \dfrac{{0.00099}}{{0.005985}}\]
\[ = \dfrac{{990}}{{5985}}\]
\[ = \dfrac{{22}}{{133}} = 0.165\]
Therefore, the probability of a person having disease being tested positive is \[0.165\].

Note: Apart from this problem, if we are given with probability data then always try to find out complementary data. Bayes theorem also called as Bayes rule is useful in finding conditional probability.