Question
Suppose that the reliability of a COVID test is
specified as follows: Of people having COVID, 90% of the test detect the disease but 10% go undetected. Of people free of COVID, 99% of the test are judged COVID–ive but 1% are diagnosed as showing COVID+ive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID test, and the pathologist reports him/her as COVID+ive. What is the probability that the person actually has COVID?Solution
Let E denote the event that the person selected is actually having COVID and A the event that the person's COVID test is diagnosed as +ive. We need to find P(E|A). Also, E’ denotes the event that the person selected is actually not having COVID. Clearly, {E, E'} is a partition of the sample space of all people in the population. We are given that P(E) = 0.1% = 0.1/100 = 0.001 P(E') = 1 – P(E) = 0.999 P(A|E) = P(Person tested as COVID+ive given that he/she is actually having COVID) = 90% = 90/100 = 0.9 and P(A|E') = P(Person tested as COVID +ive given that he/she is actually not having COVID) = 1% = 1/100 = 0.01 Now, by Bayes' theorem P(E|A) = [P(E) × P(A|E)]/[P(E) × P(A|E) + P(E') × P(A|E')] = [0.001 × 0.9]/[0.001 × 0.9 + 0.999 × 0.01] = 90/1089 = 0.083 approx.
Statement: F ≥ G > I > E ≤ P, E = S ≥ PÂ
Conclusion: I. F ≥ P         II. G > P
Statement: Y < Z > I < Q > S = M ≤ N
Conclusions:
I. S= N
II. Q > M
Statements: P = Q = R > S > T > Z; U > R < V < W > X
Conclusions:
I. W > Z
II. R < W
III. R < X
In the question, assuming the given statements to be true, find which of the conclusion (s) among given three conclusions is /are definitely true and ...
Statements: E < F > G; H < I ≤ F; E > D
Conclusions:
I. F > D
II. H < E
III. G < DWhich of the following will be definitely false if the given expression F > G ≥ H > I ≥ J > K = M ≤ N > L ≤ O is definitely true?
Statements: Â M @ N, P @ R, P & N
Conclusions:Â Â Â Â Â a ) M @ PÂ Â Â Â Â Â Â Â Â Â Â Â Â b) R & M
...Statement: F < G; H ≥ I; H ≥ K; I > G ≥ J
Conclusion:
I. G > K
II. K > J
Statements: S = R, T ≤ U, O < J, T ≤ J, U > R
Conclusion:
I. R ≥ T
II. R < T
Statement: D < F; D ≥ E > G; I ≥ H > F
Conclusion:
I. G ≥ F
II. H ≥ D
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