Question

Two numbers are selected at random from the interval \((0,1) .\) If these values are uniformly and independently distributed, by cutting the interval at these numbers, compute the probability that the three resulting line segments can form a triangle.

Short answer

The probability is 1/4.

Step-by-step solution

Define the Random Variables

Let's define two random variables X and Y representing the points where the interval is cut. Since the points are selected uniformly, both X and Y are uniform random variables on the interval (0,1).

Understand the Condition for Forming a Triangle

According to the triangle inequality theorem, for any three lengths a, b, and c to form a triangle, they must satisfy the following conditions: a + b > c, a + c > b, and b + c > a. Given we are choosing two points X and Y in (0,1), the lengths of our three segments are min(X,Y), |X-Y|, and 1-max(X,Y). Thus, the triangle inequality conditions become: min(X,Y) + |X-Y| > 1-max(X,Y), min(X,Y) + 1-max(X,Y) > |X-Y|, and |X-Y| + 1-max(X,Y) > min(X,Y).

Formulate the Problem as a Double Integration

Since X and Y are independent and uniformly distributed, the joint probability density function of X and Y is 1 on the unit square, and 0 elsewhere. The problem can now be formulated as double integration over the region in the unit square for which the triangle inequalities are all satisfied.

Compute the Double Integral

Observing the symmetry in the problem, the probability can be computed as twice the integral from 0 to 1/2 of the integral from 0 to 1 - 2x dx dy. Upon evaluating this double integral, we get a probability of 1/4. Hence, the probability that three random segments from the unit interval can form a triangle is 1/4.