Bayes' Theorem/Examples/Arbitrary Example 2

Example of Use of Bayes' Theorem

Suppose that, in a population, $6$ out of every $1000$ people has an illness $X$.

It is known that:

if a person has $X$, there is a $92 \%$ probability that a blood test will be positive for $X$

if a person does not have $X$, there is a $0 \cdotp 5 \%$ probability that a blood test will be positive for $X$.

Let a person selected at random test positive for $X$.

What is the probability that this person actually has $X$?

Let $A$ be the event: Person tests positive for $X$.

Let $B_1$ be the event: Person has $X$.

Let $B_2$ be the event: Person does not have $X$.

We have that:

$\ds \map \Pr {B_1}$

$=$

$\ds 0 \cdotp 006$

$\ds \map \Pr {B_2}$

$=$

$\ds 0 \cdotp 994$

$\ds \condprob A {B_1}$

$=$

$\ds 0 \cdotp 92$

$\ds \condprob A {B_2}$

$=$

$\ds 0 \cdotp 005$

$\ds \leadsto \ \ $

$\ds \condprob {B_1} A$

$=$

$\ds \dfrac {\map \Pr {B_1} \condprob A {B_1} } {\map \Pr {B_1} \condprob A {B_1} + \map \Pr {B_2} \condprob A {B_2} }$

$\ds $

$=$

$\ds \dfrac {0 \cdotp 006 \times 0 \cdotp 92} {0 \cdotp 006 \times 0 \cdotp 92 + 0 \cdotp 994 \times 0 \cdotp 005}$

plugging in the numbers

$\ds $

$=$

$\ds 0 \cdotp 526$

calculation

Hence the probability that a person has $X$ is $0 \cdotp 6 \%$, but after testing positive the probability is $52 \cdotp 6 \%$.

$\blacksquare$