Statistics And Probability Question 87
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
Options:
A) $ \frac{110}{221} $
B) $ \frac{2}{223} $
C) $ \frac{110}{223} $
D) $ \frac{1}{221} $
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Answer:
Correct Answer: A
Solution:
- [a] 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.
$ \therefore P(A)=\frac{1}{1000},P(B)=\frac{999}{1000} $ $ P( \frac{E}{A} )=0.99,P( \frac{E}{B} )=0.001 $ The required probability is given by $ P( \frac{A}{E} )=\frac{P(A)\times ( \frac{E}{A} )}{P(A)\times P( \frac{E}{A} )+P(B)\times P( \frac{E}{B} )} $ $ =\frac{\frac{1}{1000}\times 0.99}{\frac{1}{100}\times 0.99+\frac{999}{1000}\times 0.01}=\frac{110}{221} $
BETA
