Day 4: Geometric Distribution II HackerRank Solution

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

Objective
In this challenge, we go further with geometric distributions. We recommend reviewing the Geometric Distribution tutorial before attempting this challenge.

Task
The probability that a machine produces a defective product is . What is the probability that the defect is found during the first inspections?

Input Format

The first line contains the respective space-separated numerator and denominator for the probability of a defect, and the second line contains the inspection we want the probability of the first defect being discovered by:

1 35

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

Output Format

Print a single line denoting the answer, rounded to a scale of decimal places (i.e., format).

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

Solution

	// github.com/RodneyShag
	import java.util.Scanner;
	public class Solution {
	public static void main(String[] args) {
	/* Save input */
	Scanner scan = new Scanner(System.in);
	int numerator = scan.nextInt();
	int denominator = scan.nextInt();
	int n = scan.nextInt();
	scan.close();
	/* Geometric Series */
	double p = (double) numerator / denominator;
	double result = 0;
	for (int i = 1; i <= 5; i++) {
	result += geometric(i, p);
	}
	System.out.format("%.3f", result);
	}
	private static double geometric (int n, double p) {
	return Math.pow(1 - p, n - 1) * p;
	}
	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;
	}
	}