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A letter has come from exclusively LONDON or CLIFTON, but on the postmark only $2$ consecutive letters ''ON'' are found to be visible. What is the probability that the letter came from LONDON?

This is a question of conditional probability. Let $A$ be the event that the letter has come from LONDON. Let $B$ be the event that consecutive letters ''ON'' are found to be visible. $A\cap B$ is the event that the letter has come from LONDON and consecutive letters ''ON'' are visible. We have to find $P(A\mid B) =\frac{P(A\cap B)}{P(B)}$.

But then i am stuck. Please help me. Thanks.

learner_avid
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3 Answers3

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There is not enough information given to answer the question, since we don't know the prior probabilities of letters arriving from London or from Clifton. A reasonable assumption might be that letters are equally likely to come from any of the people living in those two places. London has a population of roughly $10$ million; the suburb Clifton of Bristol (assuming that that's the Clifton that is meant) has a population of roughly $10{,}000$. Thus the prior probabilities are roughly $0.999$ for arriving from London and $0.001$ for arriving from Clifton.

Regarding the letters on the postmark, again not enough information is given, since we don't know how these letters came to be visible and how likely they are to be from various parts of the words. A reasonable assumption is that a pair of consecutive letters was uniformly randomly selected from all such pairs. Then the probability of a letter from London showing "ON" would be $2/5$, and the probability of a letter from Clifton showing "ON" would be $1/6$.

Under these two assumptions, we can calculate the a posteriori probability of the letter having arrived from London as

$$ P(\text{London}\mid\text{ON})=\frac{P(\text{London}\cap\text{ON})}{P(\text{ON})}=\frac{0.999\cdot\frac25}{0.999\cdot\frac25+0.001\cdot\frac16}\approx0.9996\;. $$

Thus we are now even more certain that the letter must have come from London than before.

If you want to solve this very badly posed problem the way it may have been intended, you may want to use the highly unrealistic prior probabilities $1/2$ instead.

joriki
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You wish to find the posterior probability that the word is LONDON given that the letters ON are visible.   This is: $\mathsf P(A\mid B)$.

You should be able to evaluate the conditional probability that the letters ON are visible given that the word is LONDON.   This is: $\mathsf P(B\mid A)$.

You should also be able to evaluate the conditional probability that the letters ON are visible given that the word is CLIFTON.   This is: $\mathsf P(B\mid \neg A)$.

You also need to assign the prior probabilities that the envelop is from LONDON or CLIFTON.   $\mathsf P(A), \mathsf P(\neg A)$

$$\mathsf P(A\mid B) = \frac{\mathsf P(B\mid A)\,\mathsf P(A)}{\mathsf P(B\mid A)\,\mathsf P(A)+\mathsf P(B\mid \neg A)\,\mathsf P(\neg A)}$$

Graham Kemp
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  • $P(A)=\frac{1}{2},P(\neg A)=\frac{1}{2}$ but what is $P(B \mid A)$ and $P(B \mid \neg A)$? – learner_avid Oct 04 '15 at 05:56
  • Is $P(B \mid A)=\frac{2}{5}?$$P(B \mid \neg A)=\frac{1}{6}$I am mainly confused here. – learner_avid Oct 04 '15 at 05:58
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    @learner_avid Yes. All possible two consecutive letters from LONDON are ${$LO, ON, ND, DO, ON$}$. Assuming a uniform distribution of the first letter's position, $\mathsf P(B\mid A)=2/5$. Similarly $\mathsf P(B\mid\neg A)=1/6$. – Graham Kemp Oct 04 '15 at 09:11
  • @GrahamKemp: I reasoned that since the letter came either from London or Clifton, there're only three possible 'ON's that it could have been a part of- two from 'lONdON' and one from 'cliftON', so I thought it'd end up being $\frac{2}{2+1}=\frac{2}{3}$. Can you explain where I went wrong here? – harry Apr 15 '21 at 15:05
  • Well, it depends on your assumption of how the pair of letters are selected. Mine was that the weight of the individual 'ON' are not equal when they come from words of different length. – Graham Kemp Apr 15 '21 at 22:27
  • @GrahamKemp: I see. So without this being specified in the question, the answer is ambiguous, right? – harry Apr 16 '21 at 02:02
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E1: Letter (ON) came from CLIFTON

E2 : Letter (ON) came from LONDON

E : Event that 2 consecutive times the letter (ON) is visible.

P(E1) = P(E2) = 1/2

CLIFTON can be broken down to : CL,LI,IF,FT,TO,ON = 6 pairs in which one pair contains letter ON . Therefore, $P(\frac{\text{Number of times ON is present}}{\text{letter came from CLIFTON}}) = P(\frac{E}{_{E1}}) = \frac{1}{6}$

LONDON can be broken down to : LO,ON,ND,DO,ON = 5 pairs in which two pairs contain the letter ON .

$P(\frac{\text{Number of times ON is present}}{\text{letter came from LONDON}}) = P(\frac{E}{_{E2}}) = \frac{2}{5}$

Then the Probability that the letter ON came from LONDON when Just 2 consecutive letter(ON) are visible is given = $P(\frac{E2}{E}) = \frac{P(E2) * P(\frac{E}{E2})}{{P(E2) * P(\frac{E}{E2}) + P(E1) * P(\frac{E}{E1})}} = \frac{\frac{1}{2} * \frac{2}{5}}{\frac{1}{2} * [ \frac{2}{5} + \frac{1}{6}]} $

Soudipta Dutta
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