probability of printing the intended letter twice

A printing machine has n letters say a1, a2,.. an. It is operated by impulses ,each letter produced by a different impulse. Assume that there exists a constant probablity ,p, of printing the correct letter and assume independence. One of the n impulses, chosen at random , was fed into the machine twice and both times a1 was printed. Whats the probabi(Giggle)lity that the impulse chosen was meant to print a1.

Re: probability of printing the intended letter twice

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Originally Posted by

**ice_syncer** A printing machine has n letters say a1, a2,.. an. It is operated by impulses ,each letter produced by a different impulse. Assume that there exists a constant probablity ,p, of printing the correct letter and assume independence. One of the n impulses, chosen at random , was fed into the machine twice and both times a1 was printed. Whats the probabi(Giggle)lity that the impulse chosen was meant to print a1.

given that $a_1$ was selected, the distribution for the printing is

$Pr[(a_1,a_1),(\neg a_1,a_1),(a_1, \neg a_1), (\neg a_1, \neg a_1)| a_1] = \{ p^2, (1-p)p, p(1-p), (1-p)^2 \}$

From this you can see $Pr[(a_1,a_1)|a_1]=p^2$

Re: probability of printing the intended letter twice

Re: probability of printing the intended letter twice

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Originally Posted by

**ice_syncer** so ?

Pr[printed a1 2x | a1]Pr[a1] = Pr[a1 | printed a1 2x]Pr[printed a1 2x]

Pr[a1] = $\dfrac 1 n$

Pr[printed a1 2x|a1] = $p^2$ as I showed

Pr[a1 | printed a1 2x] is the value you want to find

that leaves

Pr[printed a1 2x] = Pr[printed a1 2x|a1]Pr[a1] + Pr[printed a1 2x| !a1]Pr[!a1]

Pr[printed a1 2x] = $p^2 \cdot \dfrac 1 n$ + $(1-p)^2 \cdot \dfrac {n-1} n$

I leave it to you to digest all this and come up with your final answer.

Re: probability of printing the intended letter twice

according to that the answer should be (p*p/n)/(p*p/n + (1-p)*(1-p)*(n-1)/n)) ; but the answer given is (n-1)*p*p/(1-2*p+n*p*p). The answer doesnot simplify to the given answer.