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I'm trying to solve the following exercise:

Imagine an organization wants to draw a prize amongst its $N$ members. So each one is given a number and only one of them is the winner. To verify is they have won, they introduce their number in a machine. However, the machine can be mistaken and tell you you are the winner without being the right number and viceversa (being the right number and not telling you are the winner) with a probability of $\alpha$.

So now, a student called Susan, when introducing her number, the machine tells her that she is the winner.

  1. What is the probability that she is actually the real winner?

  2. If she then introduces her number in another machine independent from the 1st one, and again tells her she has won, what would be now the probability of being the real winner? Repeating this process with $m$ machines (independent), to which value does now the probability tend to?

My try to solve this is this. First I calculate the probability of the machine to say that I win. This involves two cases: 1st: I had the right number and the machine doesn't fail; and 2nd: I didn't have the right number but the machine went wrong. That is:

$$P(\text{machine says win})=P(\text{right No.})P(\text{machine OK | right No.})+P(\text{wrong No.})P(\text{machine fails | wrong No.})$$

Now I know that $P(\text{right No.})=1/N$ and $P(\text{wrong No.})=(N-1)/N$ and. because the machine can fail regardless of whether you got the right or wrong number, then $P(\text{machine OK | right No.})=P(\text{machine OK})=1-\alpha$, and $P(\text{machine fails | wrong No.})=P(\text{machine fails})=\alpha$.

Substituting the values I get $P(\text{machine says win})=\frac{1}{N}(1+\alpha N-2\alpha)$.

Then, using Bayes Theorem, $$P(\text{right No.|machine says win})=\frac{P(\text{right No.})P(\text{machine says win | right no.})}{P(\text{machine says win})}$$

Now, $P(\text{machine says win | right no.})=P(\text{machine OK})=1-\alpha$, so: $$P(\text{right No.|machine says win})=\frac{1-\alpha}{1+\alpha N-2\alpha}$$ with is the value asked in the 1st question.

About the 2nd question I have little idea about where to start, so any help would be good; and if anyone sees any mistakes in the 1st question it would also be helpful. Thanks!

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You did the first part correctly.

For the second part, do it the same way, but now the prior event is that all the machines say "win" so we have

Pr(all say win)= Pr(win)Pr(all machines OK) + Pr(loss)Pr(all machines bad)

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