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Gaurav Seth 7 years, 1 month ago
Let E1 --> Letters came from the word LONDON
and E2 --> Letters came from the word CLIFTON
Since the letters have to come from either of these two words
P(E1) = P(E2) = 1/2
Let A --> Two consequitive letters on the envelope are ON
If E1 occurs then the letters ON come from the word LONDON. In this word there are 6 letters in which ON occurs twice. Considering one of the ON's as one object/letter there are now 5 letters.
Therefore
P(A/E1) = 2/5
Now if E2 occurs, then the letters come form the word CLIFTON. In this there are 7 letters in which ON occurs once. Considering ON as one object/letter there are now 6 letters.
Therefore
P(A/E2) = 1/6
Baye's Theorem
(i) P(E1/A) = ( 1/2 * 2/5 ) / ( 1/2 * 2/5 + 1/2 * 1/6 )
---------> = 2/5 * 30/17
---------> = 12/17 ( ans )
(ii) P(E2/A) = ( 1/2 * 1/6 ) / ( 1/2 * 1/6 + 1/2 * 2/5 )
---------> = 1/6 * 30/17
---------> = 5/17 ( ans )
Posted by Ajay Parmar 7 years, 1 month ago
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Divya Gupta 7 years, 1 month ago
0Thank You