# 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.

 `// 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.