91Ó°ÊÓ

Let's assume the line L to be a segment of length l, for simplicity. We can extend the results to an unlimited-length line by taking limits.

Let's name the positions of the points \(X_1\), \(X_2\), and \(X_3\) as \(x_1\), \(x_2\) and \(x_3\), respectively, along the line \(L\). Since the line is now a segment, \(0 \leq x_1, x_2, x_3 \leq l\).

We can form 3! = 6 arrangements for the positions of the \(X_1\), \(X_2\), and \(X_3\) along the line segment, but since the points are indistinguishable, we are only interested in two cases: 1. \(X_2\) lies between \(X_1\) and \(X_3\): In this case, \(x_1 \leq x_2 \leq x_3\). 2. \(X_2\) lies outside \(X_1\) and \(X_3\): In this case, either \(x_2 \leq x_1 \leq x_3\) or \(x_1 \leq x_3 \leq x_2\). We want to find the probability of case 1 occurring.

Since points \(X_1\), \(X_2\), and \(X_3\) are selected at random, we can divide the line L into small equal intervals (we can consider it as continuous limits). The probability of each case is the ratio of the lengths of line segments where that case is true to the total length of the line segment L. For case 1 (\(x_1\leq x_2 \leq x_3\)): Notice that we can use the triangle formed by \((0,0)\), \((x_2, x_1)\), and \((x_2, x_3)\) in the x-y coordinate system: when moving upwards and to the right of the y=x line, the condition \(x_1 \leq x_2 \leq x_3\) is satisfied and creates a right triangle whose points are on the line y=x. The area of this region can be obtained by integrating the x-axis from 0 to l and integrating with respect to the y-axis from x to l for each x-value. \[Area_1 = \int_{0}^{l} \int_{x}^{l} dy dx\] For case 2 (\(x_2\leq x_1\leq x_3\) or \(x_1\leq x_3\leq x_2\)): Notice that when combining case 1 and case 2, the entire square area is covered, so the probability of case 2 can be calculated as the complement of case 1 within a square. \[Area_{Square} = l^2\] Once we have the values of both areas, we can then calculate the conditional probabilities of the cases.

First, let's find the area for case 1: \[Area_1 = \int_{0}^{l} \int_{x}^{l} dy dx = \int_{0}^{l} (l-x) dx = -\frac{l^2}{2} + l^2 = \frac{l^2}{2}\] Now, for case 2, we can use the complement as mentioned above: \[Area_2 = Area_{Square} - Area_1 = l^2 - \frac{l^2}{2} = \frac{l^2}{2}\] The probabilities of each case are the areas divided by the total area: \[P(case_1) = \frac{Area_1}{Area_{Square}}=\frac{l^2/2}{l^2}=\frac{1}{2}\] \[P(case_2) = \frac{Area_2}{Area_{Square}}=\frac{l^2/2}{l^2}=\frac{1}{2}\] Thus, the probability of \(X_2\) lying between \(X_1\), and \(X_3\) is 0.5 (or 50%).