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Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important You are here
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
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Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Example 23 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at May 29, 2023 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 –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