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A laboratory blood test is $\mathbf{9 9} \%$ 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?
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?
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Verified Answer
Let
$\mathrm{E}_1=$ The person selected is suffering from certain disease, $\mathrm{E}_2=$ The person selected is not suffering from certain disease.
$\mathrm{A}=$ The doctor diagnoses correctly
Now $\quad \mathrm{P}\left(\mathrm{E}_1\right)=0.1 \%=\frac{1}{1000}=0.001$
$$
\begin{aligned}
&\mathrm{P}\left(\mathrm{E}_2\right)=1-\frac{1}{1000}=\frac{999}{1000}=0.999 \\
&\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=99 \%=\frac{99}{100}=0.99
\end{aligned}
$$
and $\quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=0.005$
Now by Bayes' theorem,
$$
P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}
$$
$$
=\frac{.001 \times 0.99}{.001 \times 0.99+0.999 \times 0.005}=\frac{22}{133} .
$$
$\mathrm{E}_1=$ The person selected is suffering from certain disease, $\mathrm{E}_2=$ The person selected is not suffering from certain disease.
$\mathrm{A}=$ The doctor diagnoses correctly
Now $\quad \mathrm{P}\left(\mathrm{E}_1\right)=0.1 \%=\frac{1}{1000}=0.001$
$$
\begin{aligned}
&\mathrm{P}\left(\mathrm{E}_2\right)=1-\frac{1}{1000}=\frac{999}{1000}=0.999 \\
&\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=99 \%=\frac{99}{100}=0.99
\end{aligned}
$$
and $\quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=0.005$
Now by Bayes' theorem,
$$
P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}
$$
$$
=\frac{.001 \times 0.99}{.001 \times 0.99+0.999 \times 0.005}=\frac{22}{133} .
$$
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