Question

By examining the chest $X-$ray, the probability that $TB$ is detected when a person is actually suffering from it is $0.99$. The probability of an healthy person diagnosed to have $TB$ is $0.001$. In a certain city, $1$ in $1000$ people suffers from $TB$. A person is selected at random and is diagnosed to have $TB$. What is the probability that he/she actually has $TB$?

• A. $\frac{55}{221}$
• B. $\frac{110}{221}$
• C. $\frac{111}{221}$
• D. $\frac{56}{221}$

Answer 1

The correct option is A $\frac{55}{221}$

$P\left(E_1\right)=$ Person suffering from $TB=\frac{1}{1000}$

$P\left(E_2\right)=$ Person not suffering from $TB=\frac{999}{1000}$

Let A : Doctor diagnoses correctly

Then $\left(A/E_1\right)=$ T.B detected actually suffering

$=0.99=\frac{99}{100}$

$P\left(A/E_2\right)=$ Doctor diagnoses incorrectly

$=0.001=\frac{1}{1000}$

Required Probability of Person suffering by T.B

diagnosed correctly i.e; $P\left(E_1/A\right)$

By bases theorem :-

$P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)}$

$=\frac{\frac{1}{1000}\ast \frac{99}{100}}{\frac{1}{1000}\ast \frac{99}{100}+\frac{999}{1000}\ast \frac{1}{1000}}$

$=\frac{99}{99+99.9}$

$=\frac{99}{9(11+11.1)}=\frac{11}{22.1}$

$P\left(E_1/A\right)=\frac{110}{221}$

$\therefore$ The person suffering by 7.8 diagnosed correctly

is $\frac{110}{221}$