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Probability

Question
CBSEENMA12033773

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will Come by train, bus, scooter or by other means of transport are respectively 3 over 10 comma space 1 fifth comma space 1 over 10 space and space 2 over 5. The probability that he will be late are  1 fourth comma space 1 third space and space 1 over 12 comma space if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train ?

Solution
Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4 be the events that the doctor comes by train, bus, scooter, and other means of transport respectively.
Then               straight P left parenthesis straight T subscript 1 right parenthesis space equals space 3 over 10 comma space space straight P left parenthesis straight T subscript 2 right parenthesis space equals space 1 fifth comma space space straight P left parenthesis straight T subscript 3 right parenthesis equals space 1 over 10 space and space straight P left parenthesis straight T subscript 4 right parenthesis space equals space 2 over 5
          straight P left parenthesis straight E vertical line straight T subscript 1 right parenthesis space equals space Probability space that space the space doctor space arriving space late space comes space by space train space equals space 1 fourth
Similarly, straight P left parenthesis straight E thin space vertical line thin space straight T subscript 2 right parenthesis space equals space 1 third comma space space straight P left parenthesis straight E thin space vertical line thin space straight T subscript 3 right parenthesis space equals space 1 over 12 space and space straight P left parenthesis straight E thin space vertical line thin space straight T subscript 4 right parenthesis space equals space 0, since he is not late if he comes by other means of transport.
       Therefore, by Baye's Theorem, we have
                  straight P left parenthesis straight T subscript 1 space vertical line thin space straight E right parenthesis space equals space Probability space that space the space doctor space arriving space late space comes space by space train
                                  equals space fraction numerator straight P left parenthesis straight T subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight T subscript 1 right parenthesis over denominator straight P left parenthesis straight T subscript 1 right parenthesis space straight P left parenthesis straight E thin space vertical line space straight T subscript 1 right parenthesis space plus space straight P left parenthesis straight T subscript 2 right parenthesis thin space straight P left parenthesis straight E thin space vertical line thin space straight T subscript 2 right parenthesis space plus space straight P left parenthesis straight T subscript 3 right parenthesis thin space straight P left parenthesis straight E space vertical line space straight T subscript 3 right parenthesis space plus space straight P left parenthesis straight T subscript 4 right parenthesis thin space straight P left parenthesis straight E vertical line straight T subscript 4 right parenthesis end fraction
equals space fraction numerator begin display style 3 over 10 end style cross times begin display style 1 fourth end style over denominator begin display style 3 over 10 end style cross times begin display style 1 fourth end style plus begin display style 1 fifth end style cross times begin display style 1 third end style plus begin display style 1 over 10 end style cross times begin display style 1 over 12 end style plus begin display style 2 over 5 end style cross times 0 end fraction space equals space fraction numerator begin display style 3 over 40 end style over denominator begin display style 3 over 40 end style plus begin display style 1 over 15 end style plus begin display style 1 over 120 end style plus 0 end fraction
equals space fraction numerator begin display style 3 over 40 end style over denominator begin display style fraction numerator 9 plus 8 plus 1 over denominator 120 end fraction end style end fraction space equals fraction numerator begin display style 3 over 40 end style over denominator begin display style 18 over 120 end style end fraction space equals 3 over 40 cross times 120 over 18 space equals 1 half
therefore space space space space space the space required space probability space is space 1 half.

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