I have one problem that I'm not able to solve. Let's say we have set A of 50 integers (1 to 50). We choose randomly 30 numbers (one by one, no repeat) from set A and let's name this new set, set B. Now, let's create third set C of 6 integers (possible numbers are from set A). This is some sort of a bingo game, so whats the probability that in n-th picking from set A (where we create set B) that we hit all 6 numbers from set C.
I'm looking for a formula in terms of A, B and C where this represents size of these set and n which indicates n-th picking.
Hope I was clear enough. Thanks in advance 🙂
EDIT: I wasn't clear enough.
I'm picking 30 times from set A. My question is what's the probability that in n-th picking I'll hit all 6 numbers I choosed in set C. I understand that in first 5 picks the probability is 0.
Best Answer
I hope I've understood your question correctly. I'll use $a=|A|$ and $c=|C|$.
Disregarding order, the number of ways to get all $c$ numbers in your first $n$ numbers for $B$ is the number of ways to choose the other $n-c$ numbers from the other $a-c\;$ non-$C$ numbers available. This equals $\binom{a-c}{n-c}$.
The total number of ways to choose $n$ numbers from $a$ available is $\binom{a}{n}$.
So the probability of your $n$ numbers having all $c$ numbers is $$\binom{a-c}{n-c}\bigg/\binom{a}{n}.$$
However, it sounds like you only want the cases where the $n^{th}$ number belongs to set $C$ so that you haven't got all $c$ numbers until you choose your $n^{th}$ number.
The fraction of the $\binom{a-c}{n-c}$ ways of having all $c$ numbers that have the $n^{th}$ number belonging to set $C$ is just $\dfrac{c}{n}$ because the $n^{th}$ number is equally likely to be any of the $n$ numbers selected.
So with this additional requirement, the probability is $$\dfrac{c}{n}\binom{a-c}{n-c}\bigg/\binom{a}{n}.$$