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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 random and is diagnosed to have TB. What is the probability that he actually has TB?
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. What is the probability that he actually has TB?
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
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. What is the probability that he actually has TB?

Let \[{{E}_{1}}\] = Event that a person has TB

\[{{E}_{2}}\] = Event that a person does not have TB

And H = Event that the person is diagnosed to have TB.

So,

\[P({{E}_{1}})\text{ }=\text{ }1/1000\text{ }=\text{ }0.001,\text{ }P({{E}_{2}})\text{ }=\text{ }1\text{ }\text{ }1/1000\text{ }=\text{ }999/1000\text{ }=\text{ }0.999\]

\[P(H/{{E}_{1}})\text{ }=\text{ }0.99,\text{ }P(H/{{E}_{2}})\text{ }=\text{ }0.001\]

Now, using Baye’s theorem we have

Therefore, the required probability is \[110/221\].

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