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This is taken from a stochastic processes text.

A printing machine capable of printing any of n characters $a_1, ... , a_n$ is operated by electrical impulses, each character, in theory, being produced by a different impulse. Suppose the machine has probability $p$ of producing the character corresponding to the impulse received, independent of past behavior. If it prints the wrong character, the probabilities that any of the $(n-1)$ other characters will appear are equal.

The question is: Suppose that one of the $n$ impulses is chosen at random and fed into the machine twice, and the character $a_i$ is printed both times. What is the probability that the impulse chosen is the one designed to produce $a_i$?

I thought that the answer can be found by supposing two characters are $a_i$ with probability $p$ and the remaining $(n-2)$ characters are not each with probability $(1-p)$.

I think this would be $$nchoose 2 p^2(1-p)^n-2$$

The actual answer is $$p^2[p^2 + frac(1-p)^2n-1]^-1$$

Looking at this, it seems it might be a conditional probability, but I do not understand what A and B would be in $P(A|B)$.

I also thought that you need to consider at least two characters = $a_i$ as opposed to exactly two, but I do not see how this would lead to the actual answer.

I think there is something about the problem and the question that I do not understand. Please clarify, and please show or give hints as to obtain the actual answer.

Let $A$ be the event that impulse $i$ is chosen and $B$ the event that the first and the second printed letter is $a_i$.

Use Bayes theorem to obtain the desired probability.
$$
P(A|B) = fracA)P(A)notA)P(notA)
$$

Now, $P(A) = 1/n, P(B|A) = p^2, P(B|notA)=((1-p)/(n-1))^2, P(notA)=(n-1)/n.$

Substituting in the formula, the result is $p^2/(p^2+(1-p)^2/(n-1)).$

Note that $(1-p)/(n-1)$ is the probability that letter $a_i$ is printed given that impulse other than $i$ is selected.