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Chapter 13 Class 12 Probability

Example 18 - Suppose reliability of a HIV test is: people

Example 18 - Chapter 13 Class 12 Probability - Part 2
Example 18 - Chapter 13 Class 12 Probability - Part 3

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Transcript

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 โ€“ive 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‘ƒ(๐ธ) . ๐‘ƒ(๐บ|๐ธ))/(๐‘ƒ(๐ธ) .๐‘ƒ(๐บ|๐ธ)+๐‘ƒ(๐น) . ๐‘ƒ(๐บ|๐น)) "P(E)" = Probability that the person selected has HIV = 0.1%=0.1/100=๐ŸŽ.๐ŸŽ๐ŸŽ๐Ÿ ๐‘ท(๐‘ฎ"|" ๐‘ฌ) = Probability that the test judges HIV +ve , if the person actually has HIV = 90%=90/100=๐ŸŽ.๐Ÿ— "P(F)" = Probability that the person selected does not have HIV = 1 โ€“ "P(E)" = 1โˆ’0.001=๐ŸŽ.๐Ÿ—๐Ÿ—๐Ÿ— ๐‘ท(๐‘ฎ"|" ๐‘ญ) = Probability that the test judges HIV +ve , if the person does not have HIV = 1%=1/100=๐ŸŽ.๐ŸŽ๐Ÿ Putting values in formula, P(E|G) =(0.001 ร— 0.9)/(0.001 ร— 0.9 + 0.999 ร— 0.01) =(9 ร— ใ€–10ใ€—^( โˆ’ 4))/(9 ร— ใ€–10ใ€—^( โˆ’ 4) + 99.9 ร—ใ€– 10ใ€—^( โˆ’ 4) ) =(ใ€–10ใ€—^( โˆ’ 4) ร— 9)/(ใ€–10ใ€—^( โˆ’ 4) [9 + 99.9]) =9/108.9 =๐Ÿ—๐ŸŽ/๐Ÿ๐ŸŽ๐Ÿ–๐Ÿ— = 0.083 (approx) Therefore, required probability is 0.083

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