- Viral is present in infected person and positive tat = 0.8 Negative text : 1- 0.8= 0.2 =7 P(9) - Vine is not present in a person and bat is positive : 5 10 = 0.05 P (3 ) - negative : 1-0-05 = 0.95 -> P (4 ) (1) Out of all people who tested positive , portion really infected : P(1) = 0 .9411 P ( 1 ) + P(3 ) 0.8 + 0. 05 I One is infected P s ) one taate positive " - One teats Negative. 2 = 50 = 005 1000 Two ways contingency table I I Total 0.04 0.0475 0. 6875 0.01 0. 9025 0. 9125 Total 0.05 0.95 1.08P ( I ( P ) = P ( I a P ) P ( P ) P (P ) = 0.0875 P ( In P ) = 0. 64 37 P ( I / P ) = 0.04 = 0. 4571 0.0 875 3 - False Negative Reduced to 1090 - False positive Reduced to 3% P (2 ) = 0.1 - P ( 4 ) = 1 - 0.1 = 0.9 R ( 3 ) = 0. 03 - 7 P ( H ) = 1 - 0.03 = 0.97 - If a person tested positive : =) probability 27 R ( 1 ) 0. 9 0. 9 677 P( 1 ) + P ( 3 ) 0.9 + 0.03 4 Part 1 = 0. 9411 - The text have a hogher probability in predicting the Part 2 = 0. 4571 infection when the accuracy was improved rather than Part 3 = 0. 967 7 | The scenario when the infection sate is higher since The corresponding probabilities of doing so yields 0 9627 and 0:4571, respectively , at compared to the baseline which is 0. 9411 . Explanation : In the scenario above , improving the accuracy both seduces the error - the file negative sate and fake positive sate. Under the condition where accuracy is losing improved , the probability of committing type I and It error are both reduced which are favorable in every tats