Question 2 - Case Based Questions (MCQ) - Chapter 13 Class 12 Probability

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 × 1/100 = 0.001 Question 1 What is the probability of the ‘person to be tested as COVID positive’ given that ‘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 ‘person to be tested as COVID positive’ given that ‘he is actually not having COVID’? (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 ‘person is actually not having COVID? (a) 0.998 (b) 0.999 (c) 0.001 (d) 0.111 P(person not having COVID) = 1 – P(person having COVID) P(F) = 1 – P(E) = 1− 0.001 = 0.999 So, the correct answer is (b) Question 4 What is the probability that the ‘person is actually having COVID given that ‘he is tested as COVID positive’? (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) = (𝑃(𝐸). 𝑃(𝐺|𝐸))/(𝑃(𝐸). 𝑃(𝐺|𝐸)+𝑃(𝐹). 𝑃(𝐺|𝐹)) "P(E)" i.e. Probability that the person has COVID P(E) = 0.1%=𝟎.𝟎𝟎𝟏 P(G|E) i.e. P(tested COVID positive | has COVID) This is calculated in Question 1 𝑃("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 𝑃("G|" 𝐹) = 0.01 Putting values in formula, P(E|G) = (0.001 × 0.9)/(0.001 × 0.9 + 0.999 × 0.01) = 0.0009/(0.0009 + 0.0099) = (𝟎.𝟎𝟎𝟎𝟗)/(𝟎.𝟎𝟏𝟎𝟖𝟗) = (9/1000)/(1089/100000) = (9 × 10)/1089 = 0.083 ( approx.) So, the correct answer is (c) Question 5 What is the probability that the ‘person selected will be diagnosed as COVID positive’? (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) × P(tested COVID positive | does not have COVID) + P(has covid) × P(tested COVID positive | has COVID) = P(F). P(G|F) + P(E). P(G|E) = 0.001 × 0.9 + 0.999 × 0.01 = 0.0009 + 0.0099 = 0.01089 So, the correct answer is (b)