# The task is to figure out the % probality,to the nearest percent, t...

## Question

The task is to figure out the % probality,to the nearest percent, that a triangle can be formed after breaking a piece of spaghetti in 2 random places along its length.To make it more challenging, you are allowed to nibble on one of the pieces before trying to make triangle.
Input formatAn int representing the maximum percent of the chosen piece that can e nibbled away
Output Format
An int is an integer from 0 to 100,the output is an integer from 0 to 100.
Sample input 15
Sample Output 75

## Solution Preview

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import java.util.Scanner;

/*
* Class to perform Monte Carlo analysis of
* breaking a length in 3 to make a triangle
*/
public class MonteCarloTriangle {

// Number of Monte Carlo simulations to run
final static int NUM_TRIALS = 100000;

/*
* Main method to run program
*/
public static void main(String[] args) {
// Use console input to get percent of spaghetti which can be nibbled away
Scanner s = new Scanner(System.in);
System.out.print("Enter maximum percentage of piece which can be nibbled away: ");
int nibblePercent = s.nextInt();
s.close();
// Store the number of successful outcomes
int successes = 0;
for (int i = 0; i < NUM_TRIALS; i++) {
// Simulate triangle through Monte Carlo
if (simulateTriangleMonteCarlo(nibblePercent))
successes++;
}
// Calculate percent of outcomes successful
long percent = Math.round(successes * 100.0 / NUM_TRIALS);
// Print percent to console
System.out.println(percent);
}

/*
* Method to simulate forming triangle through Monte Carlo
*/
public static boolean simulateTriangleMonteCarlo(int nibblePercent) {
// Randomly generate three side lengths by splitting
// lengths away from a total length of 1.0
double a = Math.random();
double b = (1.0 - a) * Math.random...

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