probability of choosing numbers I have 5 numbers [1, 2, 3, 4, 5]
I'm choosing 2 numbers with replacements e.g. (2, 4) or (2, 2), and always taking the lowest one (2).
What is the probability of taking each number? What if I choose 3 numbers to start with?
 A: With small numbers, a carefully done calculation will give us the answer.  With larger numbers, "theory" may be helpful, but it is all too easy to make a mistake, so a general formula should be tested against careful computations with small numbers.
I will look at the following problem.  We are drawing, with replacement, $6$ times, from the collection $\{1,2,3,\dots,40\}$ of numbers. On each draw, each of the numbers is equally likely.  Let MIN be the smallest number that we get.  We want the probability that MIN is $15$.  If you understand the calculation that follows, you will see how to solve the problem in general.  I have chosen the concrete numbers $40$, $6$, and $15$ in the hope that they will make the argument more understandable than if I had used symbols.
Let us record the results of our draws as a string of $6$ numbers.  The number of such strings is $40^6$, and all of these are equally likely.
Now how many strings have the property that their smallest number is $15$?  We will use a trick.
We first ask: how many strings have each number $\ge 15$?  There are $26$ numbers from $1$ to $40$ that are $\ge 15$, so the number of strings with all entries $\ge 15$ is $26^6$.
How many strings have each number $\ge 16$? Clearly, by the same argument, there are $25^6$ of these.
Now let $A$ be the collection of strings with all entries $\ge 15$, and $B$ be the collection of strings with all entries $\ge 16$.  Clearly, $B$ is a subset of $A$.  The strings that are in $A$ but not in $B$ are the strings with all entries $\ge 15$, but not all entries $\ge 16$, so at least one entry equal to $15$. So these are precisely the strings with MIN equal to $15$.
We conclude that there are 
$$26^6-25^6$$
strings with MIN equal to $15$, and therefore our probability is
$$\frac{26^6-25^6}{40^6}$$
We could rewrite the argument in more probabilistic language.  The probability all entries are $\ge 15$ is $(26/40)^6$.  The probability they are all $\ge 16$ is $(25/40)^6$.  To find the probability that MIN is $15$, subtract.
I prefer the first, "counting" approach because it seems more concrete. A matter of taste.
A: Hint: The process looks sort of like this: suppose we are looking at the number 1. Then we can pick at least 1 1 in $1*5 + 5*1 - 1$ different ways, and in each of them the number chosen is 1. Now go for 2... and 3...
A: 25 total possibilities of picking 2 numbers from $\left[1,\ldots,5\right]$:
Out of those 25 possibilities, which we'll represent as pairs $(x,y)$, we see that there are 9 ways of having either $(1,x)$ or $(x,1)$, without counting $(1,1)$ twice. Thus we'll have $\frac{9}{25}$ probability of it being 1. Similarly we can get the rest of the values by making a chart, with 1 through 5 on the left and 1 through 5 on top. So, Probability of it being 2 is $\frac{7}{25}$. Probability of it being 3 is $\frac{5}{25}$. Probability of it being 4 is $\frac{3}{25}$. Probability of it being 5 is $\frac{1}{25}$. 
