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  • 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

By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of a healthy person diagnosed with 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 actually has TB?

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Solution

Let events E1, E2, E3 be the following events:

E1 - the event that person has TB and E2 - the event that the person does not have TB

Therefore, Total persons = 1000

Therefore, P\left(E_{1}\right)=\frac{1}{1000}=0.001, P\left(E_{2}\right)=\frac{1000-1}{1000}=\frac{999}{1000}=0.999

Let E be the event that the person is diagnosed with TB

\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)$ is the probability that $\mathrm{TB}  is detected when a person is actually suffering

\\\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)$ the probability of an healthy person diagnosed to have TB \\Therefore, $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=0.99$ and $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=0.001
To find- the probability that the person actually has TB
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
\therefore P(A \mid B)=\frac{P(A) P(B \mid A)}{P(B)}
\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right) is the probability that a person actually has TB

\begin{aligned} &\therefore P\left(E_{1} \mid E\right)=\frac{P\left(E_{1}\right) \times P\left(E_{1} \mid E\right)}{P\left(E_{1}\right) \times P\left(E_{1} \mid E\right)+P\left(E_{2}\right) \times P\left(E_{2} \mid E\right)}\\ &=\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001}\\ &=\frac{\frac{1}{1000} \times \frac{99}{100}}{\frac{1}{1000} \times \frac{99}{100}+\frac{999}{1000} \times \frac{1}{1000}}\\ &=\frac{\frac{99}{1000 \times 100}}{\frac{990+999}{1000 \times 1000}}\\ &=\frac{990}{990+999}\\ &=\frac{990}{1989}\\ &=\frac{110}{221} \quad \text { [Dividing both numerator and denominator by } \left.9\right] \end{aligned}

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