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

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%..

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.

// 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; }

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