There is a set of numbers {1,2,3,4,5,6,7,8,9,10,11,12,13}. I am trying to find out how many subsets I can form that have 3 numbers in them and also the amount of subsets that will have a number 1 in them. Also, if I was to randomly generate a single outcome, what is the probability it will have a 1 in it if all outcomes have equal chance of occurring?
I know that the probability can be calculated from (Number of subsets that have 1 in them)/(Total number of permutations)
I figured out the Total total number of permutations = 13!/(13-3)! = 1716
But I can't figure out how to determine how many subsets will have a 1 in them. In addition to this, is there a way to generalise the method? For example, what if I would like to find out the number of subsets that have a number 1 and a number 2 in them for the same n and k parameters?
It looks like instead of subsets you mean three element permutations that include $1$. You can count those by choosing first the location of the $1$, which gives three choices, then the first other element, $12$, and the second other element, $11$ for a total of $3 \cdot 12 \cdot 11=396$. You are correct that the chance a random permutation includes a $1$ is $\frac {396}{1716}=\frac 3{13}$ This is not surprising as you are choosing three elements from thirteen.
To have both $1$ and $2$ you have three places to put the $1$, two places for the $2$, and $11$ choices for the third number.