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Probability

Question
CBSEENMA12033755

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Solution

Let E1, E2, E be the events
E1 : ‘the person has the disease’
E: ‘the person is healthy’,
E : ‘test is positive’,
        therefore space space space space straight P left parenthesis straight E subscript 1 right parenthesis space equals space 0.1 space equals space 1 over 10 space space and space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 1 minus 1 over 10 equals 9 over 10
                straight P left parenthesis straight E space vertical line space straight E subscript 1 right parenthesis space equals space 99 over 100 space and space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 0.005 space equals space 5 over 1000
Required probability = straight P left parenthesis straight E subscript 1 space vertical line space straight E right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E space vertical line space straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E space vertical line space straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis space straight P left parenthesis straight E space vertical line space straight E subscript 2 right parenthesis end fraction
                                                                            (By Bare's Theorem)
                               equals space fraction numerator begin display style 1 over 10 end style cross times begin display style 99 over 100 end style over denominator begin display style 1 over 10 end style cross times begin display style 99 over 100 end style plus begin display style 9 over 10 end style cross times begin display style 5 over 1000 end style end fraction space equals fraction numerator begin display style 99 over 1000 end style over denominator begin display style fraction numerator 990 plus 45 over denominator 10000 end fraction end style end fraction equals space 99 over 1000 cross times 10000 over 1035 space equals 22 over 23.
                        

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