# What is the probability of having two events of individual probability $1/k$ in $k$ attempts?

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.

• You might want to learn about the binomial distribution en.wikipedia.org/wiki/Binomial_distribution Jan 10, 2018 at 8:50
• 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?
– chak
Jan 10, 2018 at 8:50
• @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. Jan 10, 2018 at 8:59

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.

• Yes, that is the question. Nice answer, got it right away! Jan 10, 2018 at 8:58

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:

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