Probability of $C_n^k$ elements which contains at least K elements I'm trying to do a simple problem.
I have a set of integers:
$$N={1, 2, 3, 4, 5, 6} $$
I want to combine this elements in a set of combinations of size $k=3$ from this set, i.e. $C_6^3$, where $n=|N|$. However I want each combined set of size 3 have at least one element of set:
$$M={1, 2}$$
What's the probability of a resulting set $C_6^3$ of $N$ has at least one element of $M$?
I believe that I want the probability of my set has 1 or more elements of $M$. Using law of total probability I need the probability of the resulting set have less than 1 elements of $M$. I know that problem has a easy solution using combinations, but I want to do using law of total probability. 
 A: While it's a lot simpler to just count the combinations that don't hold any of the elements of M (C(4,3)) and subtract it from the total number of combinations, if you really want to use something total-probability-like:
Of all sets of 3 items drawn from 1 to 6, the ones that have at least one element of M can be divided into 3 cases:


*

*The ones that have only 1 in them (and not 2). This implies drawing
the remaining two elements from the 4 items not in M, so there are
C(4,2) of these, so 6.

*The ones that have only 2 in them (and not 1). This implies, again,
drawing the remaining two elements from the 4 items not in M, so
there are C(4,2) of these, so 6.

*The ones that have both 1 and 2. Then you can draw the remaining item
from the 4 items not in M, so there are 4 of these.


Sum these numbers up and divide by the number of combinations. If we've done everything ok, then adding the cases where no items of M were included (C(4,3)=4, as indicated above) should make the total equal to all possible combinations (C(6,3)=20), and indeed this is correct.
