Question

$10$% of a certain population suffer from a serious disease. A person suspected of the disease is given two independent tests. Each test makes a correct diagnosis $90$% of the time. The probability that the person really has the illness given that both tests are positive is $\frac{k}{10}$ where $k$ is

Answer 1

Let $A$ be the event that a person has the disease and let $B_i$ be the event ith lest is $+$ ve$(i=1,2).$ Then
$P\left(A\right)=0.1,P\left(\frac{B_i}{A}\right)=0.9$
$P\left(\frac{B_i}{\overline{A}}\right)=1-P\left(\frac{B_i}{A}\right)=1-0.9=0.1$
Let $B=B_1\cap B_2$
$\therefore P\left(\frac{B}{A}\right)=P\left(\frac{B_1}{A}\right)P\left(\frac{B_2}{B}\right)=0.81$
$P\left(\frac{B}{\overline{A}}\right)=\left(0.1\right)\left(0.1\right)$
$\therefore P\left(\frac{A}{B}\right)=\frac{P\left(A\right).P\left(\frac{B}{A}\right)}{P\left(A\right).P\left(\frac{B}{A}\right)+P\left(\overline{A}\right).P\left(\frac{B}{\overline{A}}\right)}$
$=\frac{\left(0.1\right)\left(0.81\right)}{\left(0.1\right)\left(0.81\right)+\left(0.9\right)\left(0.1\right)\left(0.1\right)}=\frac{0.081}{0.081+.009}=\frac{81}{90}=\frac{9}{10}=0.9$