Write a function that generates one of 3 numbers according to given probabilities

You are given a function rand(a, b) which generates equiprobable random numbers between [a, b] inclusive. Generate 3 numbers x, y, z with probability P(x), P(y), P(z) such that P(x) + P(y) + P(z) = 1 using the given rand(a,b) function.
The idea is to utilize the equiprobable feature of the rand(a,b) provided. Let the given probabilities be in percentage form, for example P(x)=40%, P(y)=25%, P(z)=35%..

Following are the detailed steps. 
1) Generate a random number between 1 and 100. Since they are equiprobable, the probability of each number appearing is 1/100. 
2) Following are some important points to note about generated random number 'r'. 
a) 'r' is smaller than or equal to P(x) with probability P(x)/100. 
b) 'r' is greater than P(x) and smaller than or equal P(x) + P(y) with P(y)/100. 
c) 'r' is greater than P(x) + P(y) and smaller than or equal 100 (or P(x) + P(y) + P(z)) with probability P(z)/100.

C++

// C++ code for approach
// This function generates 'x' with probability px/100, 'y' with 
// probability py/100  and 'z' with probability pz/100:
// Assumption: px + py + pz = 100 where px, py and pz lie 
// between 0 to 100 
int random(int x, int y, int z, int px, int py, int pz)
{       
        // Generate a number from 1 to 100
        int r = rand(1, 100);
     
        // r is smaller than px with probability px/100
        if (r <= px)
            return x;

         // r is greater than px and smaller than or equal to px+py 
         // with probability py/100 
        if (r <= (px+py))
            return y;

         // r is greater than px+py and smaller than or equal to 100 
         // with probability pz/100 
        else
            return z;
}

// This code is contributed by ajaymakvana.

C

// This function generates 'x' with probability px/100, 'y' with 
// probability py/100  and 'z' with probability pz/100:
// Assumption: px + py + pz = 100 where px, py and pz lie 
// between 0 to 100 
int random(int x, int y, int z, int px, int py, int pz)
{       
        // Generate a number from 1 to 100
        int r = rand(1, 100);
     
        // r is smaller than px with probability px/100
        if (r <= px)
            return x;

         // r is greater than px and smaller than or equal to px+py 
         // with probability py/100 
        if (r <= (px+py))
            return y;

         // r is greater than px+py and smaller than or equal to 100 
         // with probability pz/100 
        else
            return z;
}

Java

// This function generates 'x' with probability px/100, 'y'
// with probability py/100  and 'z' with probability pz/100:
// Assumption: px + py + pz = 100 where px, py and pz lie
// between 0 to 100
static int random(int x, int y, int z, int px, int py,
                  int pz)
{
    // Generate a number from 1 to 100
    int r = (int)(Math.random() * 100);

    // r is smaller than px with probability px/100
    if (r <= px)
        return x;

    // r is greater than px and smaller than or equal to
    // px+py with probability py/100
    if (r <= (px + py))
        return y;

    // r is greater than px+py and smaller than or equal to
    // 100 with probability pz/100
    else
        return z;
}

// This code is contributed by subhammahato348.

Python3

import random

# This function generates 'x' with probability px/100, 'y' with 
# probability py/100  and 'z' with probability pz/100:
# Assumption: px + py + pz = 100 where px, py and pz lie 
# between 0 to 100 
def random(x, y, z, px, py, pz):  
    
    # Generate a number from 1 to 100
    r = random.randint(1, 100)
    
    #  r is smaller than px with probability px/100
    if (r <= px):
        return x
    
    # r is greater than px and smaller than 
    # or equal to px+py with probability py/100 
    if (r <= (px+py)):
        return y
        
    # r is greater than px+py and smaller than
    # or equal to 100 with probability pz/100 
    else:
        return z
    
# This code is contributed by rohan07

C#

// This function generates 'x' with probability px/100, 'y'
// with probability py/100  and 'z' with probability pz/100:
// Assumption: px + py + pz = 100 where px, py and pz lie
// between 0 to 100
static int random(int x, int y, int z, int px, int py,
                  int pz)
{
    // Generate a number from 1 to 100
    Random rInt = new Random();
    int r = rInt.Next(0, 100);

    // r is smaller than px with probability px/100
    if (r <= px)
        return x;

    // r is greater than px and smaller than or equal to
    // px+py with probability py/100
    if (r <= (px + py))
        return y;

    // r is greater than px+py and smaller than or equal to
    // 100 with probability pz/100
    else
        return z;
}

// This code is contributed by subhammahato348.

JavaScript

// This function generates 'x' with probability px/100, 'y' with 
// probability py/100  and 'z' with probability pz/100:
// Assumption: px + py + pz = 100 where px, py and pz lie 
// between 0 to 100 
function random(x, y, z, px, py, pz)
{       
        // Generate a number from 1 to 100
        let r = Math.floor(Math.random() * 100) + 1;
     
        // r is smaller than px with probability px/100
        if (r <= px)
            return x;

         // r is greater than px and smaller than or equal to px+py 
         // with probability py/100 
        if (r <= (px+py))
            return y;

         // r is greater than px+py and smaller than or equal to 100 
         // with probability pz/100 
        else
            return z;
}

// This code is contributed by subhammahato348.

Time Complexity: O(1)

Auxiliary Space: O(1)

This function will solve the purpose of generating 3 numbers with given three probabilities.

Create Quiz