Example 18 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Bayes theorem
Ex 13.3, 2 Important
Ex 13.3, 3
Misc 3
Ex 13.3, 4 Important
Ex 13.3, 9
Ex 13.3, 5
Ex 13.3, 13 (MCQ) Important
Example 21 Important
Ex 13.3, 6 Important
Ex 13.3, 10 Important
Example 18 Important You are here
Example 17 Important
Ex 13.3, 7
Ex 13.3, 8 Important
Example 19 Important
Ex 13.3, 11
Ex 13.3, 12 Important
Example 37 Important
Example 20 Important
Example 33 Important
Misc 12
Misc 16 Important
Misc 13 Important
Bayes theorem
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