Probability of a Broken Bar Having at Least One Piece Less Than ( frac{L}{20} ): A Geometric Approach

Introduction

The problem at hand involves a bar of length ( L ) that is broken at two random spots. We aim to find the probability that at least one piece of the bar will be less than ( frac{L}{20} ). This can be approached using the principles of geometric probability, a powerful method to solve problems involving random variables and spatial geometry.

Step-by-Step Solution

Setting Up the Problem

Imagine the bar as a segment from 0 to ( L ). Let ( X_1 ) and ( X_2 ) be the two random break points where ( 0 leq X_1 leq X_2 leq L ).

The lengths of the three pieces formed are:

Piece 1: ( X_1 ) Piece 2: ( X_2 - X_1 ) Piece 3: ( L - X_2 )

Condition for Lengths

We want to find the probability that at least one of these lengths is less than ( frac{L}{20} ).

[ P{ text{min}(X_1, X_2 - X_1, L - X_2)

Complementary Probability

Instead of calculating this directly, we can calculate the complementary probability:

[ P{ text{min}(X_1, X_2 - X_1, L - X_2) geq frac{L}{20} } ]

This means all pieces must be at least ( frac{L}{20} ).

Setting Up the Inequalities

For the three lengths to be at least ( frac{L}{20} ): ( X_1 geq frac{L}{20} ) ( X_2 - X_1 geq frac{L}{20} ) which implies ( X_2 geq X_1 frac{L}{20} ) ( L - X_2 geq frac{L}{20} ) which implies ( X_2 leq frac{19L}{20} )

Visualizing the Region

On the ( X_1 )-( X_2 ) plane, we need to consider the region defined by:

( X_1 geq frac{L}{20} ) ( X_2 geq X_1 frac{L}{20} ) ( X_2 leq frac{19L}{20} )

This region is a triangle in the unit square from 0 to ( L ) for both ( X_1 ) and ( X_2 ).

Calculating the Area

The area of the triangle can be calculated as follows:

The vertices of the triangle are: ( left( frac{L}{20}, frac{19L}{20} right) ) ( left( frac{L}{20}, frac{19L}{20} - frac{L}{20} right) left( frac{L}{20}, frac{18L}{20} right) ) ( left( frac{19L}{20}, frac{19L}{20} right) )

Using the base and height:

Base ( frac{18L}{20} frac{9L}{10} ) Height ( frac{19L}{20} - frac{18L}{20} frac{L}{20} )

The area of the triangle is:

[ text{Area} frac{1}{2} times text{base} times text{height} frac{1}{2} times frac{9L}{10} times frac{L}{20} frac{9L^2}{400} ]

Total Area

The total area of the triangle formed by all possible combinations of ( X_1 ) and ( X_2 ) is ( frac{L^2}{2} ).

Final Probability

Thus, the probability that at least one piece is less than ( frac{L}{20} ) is:

[ P{ text{min}(X_1, X_2 - X_1, L - X_2)

Conclusion

The probability that at least one piece is less than ( frac{L}{20} ) is:

[ boxed{frac{191}{200}} ]