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

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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 –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) =(𝑃(𝐸) . 𝑃(𝐺|𝐸))/(𝑃(𝐸) .𝑃(𝐺|𝐸)+𝑃(𝐹) . 𝑃(𝐺|𝐹)) "P(E)" = Probability that the person selected has HIV = 0.1%=0.1/100=𝟎.𝟎𝟎𝟏 𝑷(𝑮"|" 𝑬) = Probability that the test judges HIV +ve , if the person actually has HIV = 90%=90/100=𝟎.𝟗 "P(F)" = Probability that the person selected does not have HIV = 1 – "P(E)" = 1−0.001=𝟎.𝟗𝟗𝟗 𝑷(𝑮"|" 𝑭) = Probability that the test judges HIV +ve , if the person does not have HIV = 1%=1/100=𝟎.𝟎𝟏 Putting values in formula, P(E|G) =(0.001 × 0.9)/(0.001 × 0.9 + 0.999 × 0.01) =(9 × 〖10〗^( − 4))/(9 × 〖10〗^( − 4) + 99.9 ×〖 10〗^( − 4) ) =(〖10〗^( − 4) × 9)/(〖10〗^( − 4) [9 + 99.9]) =9/108.9 =𝟗𝟎/𝟏𝟎𝟖𝟗 = 0.083 (approx) Therefore, required probability is 0.083

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Chapter 13 Class 12 Probability (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.