Probability that Exactly One Piece of a Broken Stick is Longer Than 0.5

The problem of breaking a stick into four pieces and determining the probability that exactly one of them is longer than 0.5 is a classic example in the field of geometric probability. This article explores the solution to this problem step-by-step, guided by the principles of geometric and combinatorial analysis.

Understanding the Problem

We consider a stick of length 1 that is broken at three random points, thus creating four segments. The goal is to find the probability that exactly one of these segments has a length greater than 0.5.

Breaking the Stick

To simplify, let the three break points be denoted as X1, X2, and X3, such that 0 1 2 3 . The four segments formed will have lengths L1 X1, L2 X2 - X1, L3 X3 - X2, and L4 1 - X3. We aim to find the probability that exactly one of these segments has a length greater than 0.5.

Condition for One Segment to Exceed 0.5

For exactly one segment to be longer than 0.5, we can assume without loss of generality that L1 > 0.5 and L2 ≤ 0.5, L3 ≤ 0.5, L4 ≤ 0.5. The same logic applies if any of the other segments are longer than 0.5.

Geometric Representation

The total area of the possible break points X1, X2, X3 can be represented as a point in a unit cube where 0 1 2 3 . The volume of this region is frac16;, as the total number of ways to choose 3 break points from 1 is given by the combination binom{3}{3} 1, and the total volume of the cube is 1.

Calculating the Condition

- Assume L1 > 0.5, which means X1 > 0.5.

- Since we want L2, L3, L4 ≤ 0.5, we have:

X2 - X1 ≤ 0.5 rArr; X2 ≤ X1 0.5 X3 - X2 ≤ 0.5 rArr; X3 ≤ X2 0.5 1 - X3 ≤ 0.5 rArr; X3 ≥ 0.5

Using the condition X1 > 0.5, we restrict X1 to the interval [0.5, 1]. For X2 to satisfy X2 ≤ X1 0.5, the maximum value for X1 is 1. Therefore, X2 in [X1, X1 0.5]. Additionally, X3 in [X2 0.5, 1] must also satisfy X3 ≥ 0.5.

The final calculation involves integrating these conditions. By symmetry, the probability is the same regardless of which segment is longer than 0.5. Therefore, we multiply the volume calculated for one case by 4, corresponding to the four segments.

After integrating and calculating these volumes, the final probability that exactly one piece is longer than 0.5 is:

P{exactly one piece > 0.5} frac14;

Conclusion

The probability that exactly one of the four pieces is longer than 0.5 is frac14;. This result highlights the power of geometric and combinatorial methods in solving probability problems, especially those involving random break points.