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A letter is known to have come either from LONDON or CLIFTON . On the envelope just two consecutive letters ON are visible . What is the probability that the letter has come from

(1) LONDON

(2) CLIFTON

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 )

 

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Nice Explanation dude...
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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 )
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