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Probability

Question
CBSEENMA12033775

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA are visible. Find the probability that the letter has come from CALCUTTA.

Solution

Let E1 be the event that the letter has come from TATANAGAR and E2 be the event that it has come from CALCUTTA.
Let A denote the event that the two consecutive letters on the envelope are TA.
∴   E1 and E2 are mutually exclusive and exhaustive events
straight P left parenthesis straight E subscript 1 right parenthesis space equals space 1 half comma space space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 1 half
P(A/E1) = Probability that the consecutive letters TA visible on the envelope belong to TATANAGAR.
equals space 2 over 8 space equals 1 fourth space space space space space space space space space space space space space space space space space space space open curly brackets table row cell There space are space 8 space consecutive space letters space namely end cell row cell TA comma space AT comma space TA comma space AN comma space NA comma space AG comma space GA comma space AR end cell row cell out space of space which space 2 space cases space are space favourable end cell end table space close curly brackets
Similarly P(A/E2) = Probability that the consecutive letters TA are visible belong to CALCUTTA
equals space 1 over 7 space space space space space space space space space space space space space space space space space space space space space space space open curly brackets table row cell There space are space seven space consecutive space letters end cell row cell CA comma space AL comma space LC comma space CU comma space UT comma space TT comma space TA end cell row cell out space of space which space 1 space case space is space favourable end cell end table close curly brackets       
therefore space space space straight P left parenthesis straight E subscript 2 divided by straight A right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis. space straight P left parenthesis straight A divided by 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 A divided by straight E subscript 2 right parenthesis end fraction
                        equals space fraction numerator begin display style 1 half end style cross times begin display style 1 over 7 end style over denominator begin display style 1 half end style cross times begin display style 1 fourth end style plus begin display style 1 half end style cross times begin display style 1 over 7 end style end fraction space equals fraction numerator begin display style 1 over 7 end style over denominator begin display style 1 fourth end style plus begin display style 1 over 7 end style end fraction space equals fraction numerator begin display style 1 over 7 end style over denominator begin display style fraction numerator 7 plus 4 over denominator 28 end fraction end style end fraction space equals 1 over 7 cross times 28 over 11 space equals 4 over 11                      

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