Case Based Questions (MCQ)

Chapter 13 Class 12 Probability
Serial order wise

## (d) 0.189

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### Transcript

Question The reliability of a COVID PCR test is specified as follows: Of people having COVID, 90% of the test detects the disease but 10% goes undetected. Of people free of COVID, 99% of the test is judged COVID negative but 1% are diagnosed as showing COVID positive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID PCR test, and the pathologist reports him/her as COVID positive. Based on the above information, answer the following:Given, Of people having COVID, 90% of the test detects the disease but 10% goes undetected. Of people free of COVID, 99% of the test is judged COVID negative but 1% are diagnosed as showing COVID positive. 90% detected 10% undetected Does not have COVID 99% COVID negative 1% COVID positive Let, E : The event that person selected has COVID F : The event that person selected does not have COVID G : The event that person is tested positive Also, From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID PCR test, and the pathologist reports him/her as COVID positive. P(person has COVID) = P(E) = 0.1% = 0.1 Γ 1/100 = 0.001 Question 1 What is the probability of the βperson to be tested as COVID positiveβ given that βhe is actually having COVID? (a) 0.001 (b) 0.1 (c) 0.8 (d) 0.9Has COVID 90% detected 10% undetected P(tested COVID | has COVID) = P(G|E) = 90% = 90/100 = 0.9 So, the correct answer is (d) Question 2 What is the probability of the βperson to be tested as COVID positiveβ given that βhe is actually not having COVIDβ? (a) 0.01 (b) 0.99 (c) 0.1 (d) 0.001 Does not have COVID 99% COVID negative 1% COVID positive P(tested COVID positive| does not has COVID) = P(G|F) = 1% = 1/100 = 0.01 So, the correct answer is (a) Question 3 What is the probability that the βperson is actually not having COVID? (a) 0.998 (b) 0.999 (c) 0.001 (d) 0.111 P(person not having COVID) = 1 β P(person having COVID) P(F) = 1 β P(E) = 1β 0.001 = 0.999 So, the correct answer is (b) Question 4 What is the probability that the βperson is actually having COVID given that βhe is tested as COVID positiveβ? (a) 0.83 (b) 0.0803 (c) 0.083 (d) 0.089 We need to find Probability that the person is actually having COVID given that he is tested as COVID positive i.e. P (having covid | covid positive) i.e. P (E|G) Now, P(E|G) = (π(πΈ). π(πΊ|πΈ))/(π(πΈ). π(πΊ|πΈ)+π(πΉ). π(πΊ|πΉ)) "P(E)" i.e. Probability that the person has COVID P(E) = 0.1%=π.πππ P(G|E) i.e. P(tested COVID positive | has COVID) This is calculated in Question 1 π("G|E") = 0.9 "P(F)" i.e. Probability that the person does not have COVID This is calculated in Question 3 "P(F)" = 0.999 P(G|F) i.e. P(tested COVID positive| does not has COVID) This is calculated in Question 2 π("G|" πΉ) = 0.01 Putting values in formula, P(E|G) = (0.001 Γ 0.9)/(0.001 Γ 0.9 + 0.999 Γ 0.01) = 0.0009/(0.0009 + 0.0099) = (π.ππππ)/(π.πππππ) = (9/1000)/(1089/100000) = (9 Γ 10)/1089 = 0.083 ( approx.) So, the correct answer is (c) Question 5 What is the probability that the βperson selected will be diagnosed as COVID positiveβ? (a) 0.1089 (b) 0.01089 (c) 0.0189 (d) 0.189P(person selected will be diagnosed as COVID positive) = P (person does not have covid and is covid positive) + P(person has covid and is covid positive) = P(does not have covid) Γ P(tested COVID positive | does not have COVID) + P(has covid) Γ P(tested COVID positive | has COVID) = P(F). P(G|F) + P(E). P(G|E) = 0.001 Γ 0.9 + 0.999 Γ 0.01 = 0.0009 + 0.0099 = 0.01089 So, the correct answer is (b)