Bayes' Theorem/Examples/Arbitrary Example 2
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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 $X$?
Solution
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:
\(\ds \map \Pr {B_1}\) | \(=\) | \(\ds 0 \cdotp 006\) | ||||||||||||
\(\ds \map \Pr {B_2}\) | \(=\) | \(\ds 0 \cdotp 994\) | ||||||||||||
\(\ds \condprob A {B_1}\) | \(=\) | \(\ds 0 \cdotp 92\) | ||||||||||||
\(\ds \condprob A {B_2}\) | \(=\) | \(\ds 0 \cdotp 005\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \condprob {B_1} A\) | \(=\) | \(\ds \dfrac {\map \Pr {B_1} \condprob A {B_1} } {\map \Pr {B_1} \condprob A {B_1} + \map \Pr {B_2} \condprob A {B_2} }\) | Bayes' Theorem | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {0 \cdotp 006 \times 0 \cdotp 92} {0 \cdotp 006 \times 0 \cdotp 92 + 0 \cdotp 994 \times 0 \cdotp 005}\) | 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 $52 \cdotp 6 \%$.
$\blacksquare$
Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Bayes' theorem