to find probability of the given problem or 
I solved this question like this but still doubt is there in my mind in the part where London is mentioned.
let us take first equally likely probability that letter could be from London or Washington
now, let $p(E)$ represent the probability that on is the only word legible then we have 
$p(E)= 0.5*p(E_1)+0.5*p(E_2)$
where $p(E_1)$ represents the probability of the only word on which is legible 
so how to calculate this $p(E_1)$ and $p(E_2)$.
Or is my approach wrong all together?!
 A: This problem is massively underspecified - you need to make lots of assumptions in order to get an answer, and different assumptions produce different answers.
The first question is the prior probability. In the absence of any other information, $P(L)=P(W)=1/2$ might be a reasonable prior (although the population of London is significantly higher than that of Washington, so maybe not).
Next, how on earth are we supposed to model which letters are visible (even assuming that in each case the postmark will consist precisely of the city name)? If each letter is independently visible with some probability $p$, well, it will depend on $p$ what answer we get, and for say $p=1/2$ a lot of the reason Washington is unlikely is simply that it has more letters, so it's unlikely that so few would still be visible. Or we could assume there are always two consecutive letters visible, with each possibility being equally likely. Or we could assume that a random contiguous section of the word is visible.
Once we have guessed at what probabilities to assign to ON being all that is visible in each case, then the solution is
$$P(L\mid E)=\frac{P(E\mid L)P(L)}{P(E\mid L)P(L)+P(E\mid W)P(W)},$$
where $L$ is the event that it came from London and $E$ that only ON is visible. In the case that two consecutive letters, equally likely to be any of the consecutive pairs, are always visible, for example, we would have $P(E\mid L)=2/5$, since of the $5$ consecutive pairs, $2$ are ON, and $P(E\mid W)=1/9$. (This gives an answer of $18/23$.)
A: This problem cannot be solved without further assumptions. Such an assumption is that the mail has come from London is $\frac12.$ Another assumption is that the letters in the word on the stamp delete randomly, independently and the probability that a letter gets deleted is $\frac12$ again.
What is the probability then that given WASHINGTON the remaining letters are ON? There are $10$ letters here and the only possibility is that the first $8$ letters get deleted and the last two remain. The probability is $$P(ON\mid  WASHINGTON)=\frac 1{2^{10}}=\frac1{1024}.$$
In the case of LONDON there are two possibilities and the number of letters is only $6$. So the conditional probability sought for is
$$P(ON\mid LONDON)=\frac2{2^6}=\frac1{2^5}.$$
The probability that the mail came from London provided the letters on the envelop is 
$$P(LONDON\mid ON)=\frac{P(ON\cap LONDON)}{P(ON)}=\frac{P(ON\mid LONDON)\frac12}{P(ON)}=\frac1{2^6}\frac1{P(ON)}$$
where 
$$P(ON)=P(ON\mid LONDON)\frac12+P(ON\mid WHASHINGTON)\frac12=$$
$$=\frac{\frac1{2^6}}{\frac1{2^6}+\frac1{2^{11}}}=\frac{2^5}{1+2^5}\approx 0.97.$$
While $$P(WASHINGTON\mid ON)=\frac{2^6}{2^6+2^{11}}\approx 0.03.$$
A: You can read "ON" three times (2 of them belonging to London, and the other one to Washington).
So, the probability of being from London is 2/3.
