Obviously, to choose $n$ elements $X = \{x_1, x_2, \dotsc, x_n\}$ of a finite set $S$ so as to maximize their combined probability $P(X) = P(x_1) + P(x_2) + \dotsb + P(x_n)$, we may simply sort the elements of $S$ in descending order of probability and take the $n$ first ones.
Similarly, to choose the smallest subset $X \subset S$ that has at least some given combined probability $P(X) \ge P_\min$, we may again sort the elements of $S$ in descending order of probability and take as many elements, starting from the first, as we need to reach the desired combined probability $P_\min$.
If you desire some more complicate tradeoff between the number of chosen elements and their combined probability, you should first try to make clear what that tradeoff should be. In any case, for any preference function $f: \mathcal P(S) \to \mathbb R$ that satisfies $f(X) \ge f(X')$ whenever $|X| \le |X'|$ and $P(X) \ge P(X')$, the subset maximizing $f$ can always be found by sorting the elements of $S$ in descending order of probability and taking the first $k$ elements for some $k$; however, there may or may not be any way to determine the optimal value of $k$ except by trying all values from $0$ to $|S|$.
(To prove this, note that, for any $X \subset S$, the set $X' = \{x_1, x_2, \dotsc, x_n\}$, where $n = |X|$ and $x_1, x_2, \dotsc, x_{|S|}$ are the elements of $S$ sorted in descending order of probability, has $|X'| = |X|$ and $P(X') \ge P(X)$, and thus $f(X') \ge f(X)$.)