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 $T B$ 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

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 $T B$ 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, $\frac{110}{221}$, $\frac{2}{223}$, $\frac{110}{223}$, $\frac{1}{221}$

MARKS App · Accepted Answer

Let A denote the event that the person has TB Let $\mathrm{B}$ denote the event that the person has not TB. Let $\mathrm{E}$ denote the event that the person is diagnosed to have TB. $\therefore \mathrm{P}(\mathrm{A})=\frac{1}{1000}, \mathrm{P}(\mathrm{B})=\frac{999}{1000}$ $\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}}\right)=0.99, \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{B}}\right)=0.001$ The required probability is given by $\begin{aligned} & P\left(\frac{A}{E}\right)=\frac{P(A) \times\left(\frac{E}{A}\right)}{P(A) \times P\left(\frac{E}{A}\right)+P(B) \times P\left(\frac{E}{B}\right)} \ & =\frac{\frac{1}{1000} \times 0.99}{\frac{1}{1000} \times 0.99+\frac{999}{1000} \times 0.001} \ & =\frac{0.99}{0.99+0.001 \times 0.999}=\frac{0.99}{0.99+0.999} \ & =\frac{990}{990+999}=\frac{990}{1989}=\frac{110}{221} \end{aligned}$

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