\section*{Problem 6B - 121236B}

If \(n\) is a natural number greater than 1, then on a line segment \(A B\), points \(P_{1}, P_{2}, P_{3}, \ldots, P_{2 n-1}\) are placed in this order such that they divide the line segment \(A B\) into \(2 n\) equal parts.

a) Give (as a function of \(n\)) the probability that two points \(P_{k}, P_{m}\) chosen from the points \(P_{1}, P_{2}, P_{3}, \ldots, P_{2 n-1}\) with \(0 < k < m < 2 n\) divide the line segment \(A B\) in such a way that a triangle can be constructed from the three segments \(A P_{k}, P_{k} P_{m}, P_{m} B\).

b) Investigate whether this probability converges to a limit as \(n \rightarrow \infty\), and determine this limit if it exists.

Note: The probability sought in a) is defined as follows: Each selection of two points \(P_{k}, P_{m}\) with \(0 < k < m < 2 n\) is referred to as a "case."

A "case" is called a "favorable case" if \(P_{k}\) and \(P_{m}\) are chosen such that a triangle can be formed from the segments \(A P_{k}, P_{k} P_{m}\), and \(P_{m} B\).

If \(z\) is the total number of possible "cases" and \(z_{1}\) is the number of "favorable cases," then the probability is defined as the quotient \(\frac{z_{1}}{z}\).