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.

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

This article is contributed by **Harsh Agarwal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above