# Example 18 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

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Chapter 13 Class 12 Probability (Term 2)

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Last updated at Feb. 15, 2020 by Teachoo

Example 18 Suppose that the reliability of a HIV test is specified as follows: Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV โive but 1% are diagnosed as showing HIV +ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV + ive. What is the probability that the person actually has HIV?Let E : person selected has HIV F : person selected does not have HIV G: test judges HIV +ve We need to find the Probability that the person selected actually has HIV, if the test judges HIV +ve i.e. P(E"|"G) P(E|G) =(๐(๐ธ) . ๐(๐บ|๐ธ))/(๐(๐ธ) .๐(๐บ|๐ธ)+๐(๐น) . ๐(๐บ|๐น)) "P(E)" = Probability that the person selected has HIV = 0.1%=0.1/100=๐.๐๐๐ ๐ท(๐ฎ"|" ๐ฌ) = Probability that the test judges HIV +ve , if the person actually has HIV = 90%=90/100=๐.๐ "P(F)" = Probability that the person selected does not have HIV = 1 โ "P(E)" = 1โ0.001=๐.๐๐๐ ๐ท(๐ฎ"|" ๐ญ) = Probability that the test judges HIV +ve , if the person does not have HIV = 1%=1/100=๐.๐๐ Putting values in formula, P(E|G) =(0.001 ร 0.9)/(0.001 ร 0.9 + 0.999 ร 0.01) =(9 ร ใ10ใ^( โ 4))/(9 ร ใ10ใ^( โ 4) + 99.9 รใ 10ใ^( โ 4) ) =(ใ10ใ^( โ 4) ร 9)/(ใ10ใ^( โ 4) [9 + 99.9]) =9/108.9 =๐๐/๐๐๐๐ = 0.083 (approx) Therefore, required probability is 0.083