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Data Interview Question

Chance of Creating a Triangle from a Divided Stick

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Solution & Explanation

The problem at hand is a classic probability puzzle often referred to as the "Broken Stick Problem." The main goal is to determine the probability that three segments, formed by breaking a stick at two random points, can form a triangle.

Problem Breakdown

  • Initial Setup:
    • You have a stick of unit length (1 unit).
    • This stick is broken at two random points along its length, creating three segments.
  • Objective:
    • Determine the probability that these three segments can form a triangle.

Triangle Inequality Theorem

To form a triangle from three segments, the triangle inequality theorem must hold:

  • The sum of the lengths of any two segments must be greater than the length of the third segment.

Mathematical Representation

  • Let the two break points be represented by random variables XX and YY such that 0<X<Y<10 < X < Y < 1.
  • This results in three segments:
    • Segment 1: XX
    • Segment 2: YXY - X
    • Segment 3: 1Y1 - Y

Applying the Triangle Inequality

  1. Condition 1:

    • X+(YX)>1YX + (Y - X) > 1 - Y
    • Simplifies to: Y>0.5Y > 0.5
  2. Condition 2:

    • X+(1Y)>YXX + (1 - Y) > Y - X
    • Simplifies to: X<0.5X < 0.5
  3. Condition 3:

    • (YX)+(1Y)>X(Y - X) + (1 - Y) > X
    • This condition is always true because 0<X<Y<10 < X < Y < 1.

Geometric Interpretation

  • The conditions Y>0.5Y > 0.5 and X<0.5X < 0.5 define a triangular region in the unit square 0<X<Y<10 < X < Y < 1.
  • The triangle is formed by the intersection of the lines Y=0.5Y = 0.5 and X=0.5X = 0.5 within the unit square.
  • This triangular region occupies one-fourth (14\frac{1}{4}) of the total area of the unit square.

Conclusion

  • Probability:
    • The probability that the three segments can form a triangle is 14\frac{1}{4} or 0.25.

Simulation

To verify this probability, a simulation can be performed in Python:

import numpy as np

# Number of simulations
N = 1000000

# Generate random break points
X = np.random.uniform(0, 1, N)
Y = np.random.uniform(0, 1, N)

# Ensure X < Y
X, Y = np.minimum(X, Y), np.maximum(X, Y)

# Compute the number of successful trials
count = np.sum((X < 0.5) & (Y > 0.5))

# Calculate the probability
probability = count / N
print("Estimated Probability:", probability)

The simulation should yield a probability close to 0.25, affirming the theoretical solution.