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My question came up while reading this other question.

We have an event with the probability $p=\frac{1}{k}$. This means that we expect $k$ attempts until we see the event.

What would the probability of two of these events happening in $k$ attempts be?

Note: I haven't included an attempt at a solution here. My intuition about probability is really bad, mostly because I never liked the subject of probability and statistics, but I need to get into it for future studies, so here I am. I don't know where to begin with this, so help would be appriciated.

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    $\begingroup$ You might want to learn about the binomial distribution en.wikipedia.org/wiki/Binomial_distribution $\endgroup$
    – clark
    Jan 10, 2018 at 8:50
  • $\begingroup$ If i ask you the probability of having this event at the first and at the second attempts and then to not have it anymore for the remaing $k-2$ attempts what it would be? $\endgroup$
    – chak
    Jan 10, 2018 at 8:50
  • $\begingroup$ @clark Thanks for pointing me to that! You could have made an answer and included that link and gotten som rep for it, instead of just commenting. $\endgroup$
    – Skillzore
    Jan 10, 2018 at 8:59

2 Answers 2

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So, the question is (is it?) that we perform $k$ independent experiments and would like to know the probability that there are exactly two successes assuming that the probability of success is $\frac1k$ in every cases.

Since the successes may take place any two positions in the sequence of experiments we may say that there are $k\choose2$ different possibilities of the same probability. So, the probability sought for is

$${k\choose2}\left(\frac1{k}\right)^2\left(1-\frac1k\right)^{k-2}.$$

This is the case because at the choosen positions the experimnet has to be successful and at the other positions it has to be unsuccessful.

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  • $\begingroup$ Yes, that is the question. Nice answer, got it right away! $\endgroup$
    – Skillzore
    Jan 10, 2018 at 8:58
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The probability of observing m events each of probability p in n experiments is:

$$Pr(X_n=m)=\binom{n}{m}p^m(1-p)^{n-m}$$

what we want is a special case in which $n=k,m=2,p={1\over k}$ therefore:

$$Pr=\binom{k}{2}{1\over {k^2}}({{k-1}\over{k}})^{k-2}$$

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