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(B) Let $A_1$ be the event that the letter came from $LONDON$ and $A_2$ be the event that the letter came from $CLIFTON$. Assuming the probability of

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$\frac{12}{17}$

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(B) Let $A_1$ be the event that the letter came from $LONDON$ and $A_2$ be the event that the letter came from $CLIFTON$. Assuming the probability of selecting either city is equal,$P(A_1) = P(A_2) = \frac{1}{2}$.In $LONDON$ (length $6$),there are $5$ pairs of consecutive letters: $(LO, ON, ND, DO, ON)$. The number of $ON$ pairs is $2$. Thus,the probability of getting $ON$ given $A_1$ is $P(E|A_1) = \frac{2}{5}$.In $CLIFTON$ (length $7$),there are $6$ pairs of consecutive letters: $(CL, LI, IF, FT, TO, ON)$. The number of $ON$ pairs is $1$. Thus,the probability of getting $ON$ given $A_2$ is $P(E|A_2) = \frac{1}{6}$.Using Bayes' Theorem,the probability that it came from $LONDON$ given $ON$ is legible is:$P(A_1|E) = \frac{P(A_1)P(E|A_1)}{P(A_1)P(E|A_1) + P(A_2)P(E|A_2)}$$P(A_1|E) = \frac{\frac{1}{2} 	imes \frac{2}{5}}{\frac{1}{2} 	imes \frac{2}{5} + \frac{1}{2} 	imes \frac{1}{6}}$$P(A_1|E) = \frac{\frac{2}{5}}{\frac{2}{5} + \frac{1}{6}} = \frac{\frac{2}{5}}{\frac{12+5}{30}} = \frac{2}{5} 	imes \frac{30}{17} = \frac{12}{17}$.

$A$ letter is known to have come either from $LONDON$ or $CLIFTON$; on the postmark only the two consecutive letters $ON$ are legible. The probability that it came from $LONDON$ is

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