Day 4: Binomial Distribution II HackerRank Solution

Day 4: Binomial Distribution II HackerRank Solution
Source : https://www.hackerrank.com/challenges/s10-binomial-distribution-2

Objective
In this challenge, we go further with binomial distributions. We recommend reviewing the previous challenge's Tutorial before attempting this problem.

Task
A manufacturer of metal pistons finds that, on average, of the pistons they manufacture are rejected because they are incorrectly sized. What is the probability that a batch of pistons will contain:

1. No more than rejects?
2. At least rejects?

Input Format

A single line containing the following values (denoting the respective percentage of defective pistons and the size of the current batch of pistons):

12 10

If you do not wish to read this information from stdin, you can hard-code it into your program.

Output Format

Print the answer to each question on its own line:

1. The first line should contain the probability that a batch of pistons will contain no more than rejects.
2. The second line should contain the probability that a batch of pistons will contain at least rejects.

Round both of your answers to a scale of decimal places (i.e., format).

Source : https://www.hackerrank.com/challenges/s10-binomial-distribution-2

Solution

	// github.com/RodneyShag
	public class Solution {
	public static void main(String[] args) {
	/* Values given in problem statement */
	double p = 0.12;
	int n = 10;
	/* "No more than 2 rejects" */
	double result = 0;
	for (int x = 0; x <= 2; x++) {
	result += binomial(n, x, p);
	}
	System.out.format("%.3f%n", result);
	/* "At least 2 rejects" */
	result = 1 - binomial(n, 0, p) - binomial(n, 1, p);
	System.out.format("%.3f%n", result);
	}
	private static Double binomial(int n, int x, double p) {
	if (p < 0 || p > 1 || n < 0 || x < 0 || x > n) {
	return null;
	}
	return combinations(n, x) * Math.pow(p, x) * Math.pow(1 - p, n - x);
	}
	private static Long combinations(int n, int x) {
	if (n < 0 || x < 0 || x > n) {
	return null;
	}
	return factorial(n) / (factorial(x) * factorial(n - x));
	}
	private static Long factorial(int n) {
	if (n < 0) {
	return null;
	}
	long result = 1;
	while (n > 0) {
	result *= n--;
	}
	return result;
	}
	}