Question
By examining the chest X-ray, the probability that TB is detected when a person is actually suffering 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. Then, the probability that the person actually has TB is :
A.
$$\frac{{110}}{{221}}$$
B.
$$\frac{2}{{223}}$$
C.
$$\frac{{110}}{{223}}$$
D.
$$\frac{1}{{221}}$$
Answer :
$$\frac{{110}}{{221}}$$
Solution :
Let $$A$$ denote the event that the person has TB.
Let $$B$$ denote the event that the person has not TB.
Let $$E$$ denote the event that the person is diagnosed to have TB.
$$\eqalign{
& \therefore \,P\left( A \right) = \frac{1}{{1000}},\,P\left( B \right) = \frac{{999}}{{1000}} \cr
& P\left( {\frac{E}{A}} \right) = 0.99,\,P\left( {\frac{E}{B}} \right) = 0.001 \cr
& {\text{The required probability is given by}} \cr
& P\left( {\frac{A}{E}} \right) = \frac{{P\left( A \right) \times \left( {\frac{E}{A}} \right)}}{{P\left( A \right) \times P\left( {\frac{E}{A}} \right) + P\left( B \right) \times \left( {\frac{E}{B}} \right)}} \cr
& = \frac{{\frac{1}{{1000}} \times 0.99}}{{\frac{1}{{1000}} \times 0.99 + \frac{{999}}{{1000}} \times 0.001}} \cr
& = \frac{{110}}{{221}} \cr} $$