# A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, 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 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?

$\begin{array}{1 1} 0.1654 \\ 0.1982 \\ 0.1827 \\ 0.1346 \end{array}$

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Let $E_1$: event that the person has disease $\rightarrow$ P($E_1$) = 0.1% = 0.0001.
Let $E_2$: event that the person does not have disease $\rightarrow$ P($E_2$) = 1 - 0.1% = 1 - 0.0001 = 0.999.
Let A: be the event that the test result is positive:
Probability that the result is positive given the person has disease = P(A|$E_1$) = 99% = 0.99
Probability that the result is positive given the person does not have disease = P(A|$E_2$) = 0.5% = 0.0005.
Since the two events are mutually exclusive and exhaustive, we can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
Using Baye's theorem, the probability that the person has disease given that his result is positive = P ($E_1$|A) =$\;\large \frac{0.0001 \;\times\; 0.99}{0.001 \;\times\; 0.99 + 0.9999\; \times\; 0.0005}$ = $\;\large\frac{0.00099}{0.005985}$= 0.1654.