How to expand the range of random values ​​from 0 to 1 inclusive?

Question

Mathematics

Hello, gentlemen programmers, I was offered to solve such a problem - in the interval [0; 1] a and b are randomly selected, so that x^2 + a*x - 2*b = 0. The value k = max(x1,x2) . Find the probability that k > 1.

I have implemented a C program that finds this probability:

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

using namespace std;
long double max(long double x1, long double x2)
{
  return x1 > x2 ? x1 : x2;
}
int main()
{
  setlocale(LC_ALL, "");
  srand(time(NULL));

  int n, count, N = 0;
  long double a, b, discr, x1, x2;

  cout << "Введите количество экспериментов:"; cin >> n;

  for(int i = 1; i <= n; i++)
  {
    a = double(rand() % 1000000000000) / double(1000000000000);
    b = double(rand() % 1000000000000) / double(1000000000000);
    discr = a*a - 8*b;
    x1 = (-a + pow(discr, 1.0/2.0))/2.0;
    x2 = (-a - pow(discr, 1.0/2.0))/2.0;

    if(discr < 0)
      continue;
    else if( max(x1,x2) > 1 )
    {
      cout << "При a = " << a << " и b = " << b << " x1 = " << x1 << " x2 = " << x2 << " max = " << max(x1,x2) << endl;
      ++count;
    }
  }

   printf("\nP(A) = %.10f", float(count) / float(n));

return 0;
}

but even with a billion experiments it gives a probability of 0, there are suspicions that the problem is in the range of random numbers, how to expand it?

Answer 1

To begin with, it is worth checking whether there are any x values ​​\u200b\u200bthat are greater than one.
D = a ² + 8×b, D ∈ [0, 9]
x ₁ = (−a + √D) / 2, x ₁ ∈ [-1, √2]
x ₂ = (−a − √D) / 2, x ₁ ∈ [-2, 0]
There are required values. And they are achieved only at one of the roots. So it makes sense to write a program and determine the probability using the Monte Carlo method.

const monteCarlo = (tryCount) => {
  let positive = 0;
  for (let i = 0; i < tryCount; i += 1) {
    const a = Math.random();
    const b = Math.random();
    const D = a * a + 8 * b;
    const x = (-a + Math.sqrt(D)) / 2;
    if (x > 1) {
      positive += 1;
    }
  }
  return(positive / tryCount);
}
monteCarlo(1000000); // 0.249374
monteCarlo(10000000); // 0.2500279

Now let's solve it analytically. We need to find the ratio of the areas of the square (0,0)-(1,1) and that part of it on which x > 1.
To do this, we solve the equation
(−a + √(a ² + 8×b)) / 2 > 1
−a + √(a ² + 8×b) > 2
√(a ² + 8×b) > a + 2
a ² + 8×b > a ² + 4×a + 4
8×b > 4×a + 4
b > (a + 1) / 2
If we draw a line graph b = (a + 1) / 2, we get that it separates 1/4 of the area of ​​the square. Thus, given the uniform distribution of a and b, the desired probability is 0.25.

Answer 2

discr = a*a - 8*b;

need +