Is there a more general expression for the probability of any subset of a subset being contained within another subset?

Question

Let:

$\phantom{2}N = \{ 1 .. n \}$

$\phantom{2}J \subset N$, where $\vert J \vert = j$

$\phantom{2}K \subset N$, where $\vert K \vert = k$ and $k \le j$

$\phantom{2}L$ any $\subset\ K$, where $\vert L \vert = l$

For example: $n = 45$, $j = 7$, $k = 6$, $l = 5$

What is the probability that any $L \subset J$?

N.B. the 'any' is important, implying that J K are specific instances, but L is all instances (I'm not sure how to notate this, advice welcome).

So the case $k = j$ is well known:

$\phantom{2}$ $P = \binom{j}{l}.\binom{n-j}{j-l}/\binom{n}{j}$

But I've had no luck finding any results or discussion on the more general case where $k \le j$.

For context this is basically a lottery problem. j is the number of numbers picked by an entrant, $l=k$ is the case of winning the main prize, and $l<k$ are the cases for other lesser prizes. Many lotteries limit the number of picks to the number drawn, but there are those which will allow a greater number of picks.

Answer 1

Doing this as a lottery problem you get a hypergeometric probability that with $n$ possibilities you get an overlap size exactly $l$ between a random set size $j$ and a random set size $k$ $($in fact one of the random sets can be fixed and it does not change the probability, but both cannot as this would make the answer $0$ or $1)$. This is $$\frac{{k\choose l}{n-k\choose j-l}}{{n\choose j}}$$ and with $n = 45$, $j = 7$, $k = 6$, $l = 5$ you get $\frac{6\times 741}{45379620}$ which is just under $10^{-4}$