HackerRank Day 4: Binomial Distribution II | 10 Days of Statistics problem solution

In this Hackerrank Day 4: Binomial Distribution II 10 Days of Statistics problem A manufacturer of metal pistons finds that, on average, 12% of the pistons they manufacture are rejected because they are incorrectly sized. What is the probability that a batch of 10 pistons will contain:

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

Problem solution in Python programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
def fact(n):
    return 1 if n == 0 else n*fact(n-1)

def comb(n, x):
    return fact(n) / (fact(x) * fact(n-x))

def b(x, n, p):
    return comb(n, x) * p**x * (1-p)**(n-x)

p, n = list(map(int, input().split(" ")))
print(round(sum([b(i, n, p/100) for i in range(3)]), 3))
print(round(sum([b(i, n, p/100) for i in range(2, n+1)]), 3))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    private static int fact(int n) {
        if(n <= 1) {
           return 1;
        } else {
            return n * fact(n-1);
        }
    }
    
    private static double bio(double a, double b, int end, int n) {
        double sum = 0;
        for(int i = 0;i <= end;i++) {
            int m = fact(n) / (fact(n-i) * fact(i));
            sum += Math.pow(a, n-i) * Math.pow(b, i) * m;
        }
        return sum;
    }
    
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scan = new Scanner(System.in);
        double percent = scan.nextInt() / 100.0;
        double pistons = scan.nextInt() / 100.0;
        
        System.out.println(String.format("%.3f", bio(1-percent, percent, 2, 10)));
        System.out.println(String.format("%.3f", bio(percent, 1-percent, 8, 10)));
    }
}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
long long int gt(int a){
    if (a<=1){
        return 1;
    }
    else{
        return a*gt(a-1);
    }
}
long long int th(int a,int b){
    return gt(a)/(gt(b)*gt(a-b));
}
int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    double a,b;
    cin >> a >> b;
    double p=a/100;
    double r1,r2;
    r1=th(b,0)*pow(p,0)*pow(1-p,b-0)+th(b,1)*pow(p,1)*pow(1-p,b-1)+th(b,2)*pow(p,2)*pow(1-p,b-2);
    r2=1-(th(b,0)*pow(p,0)*pow(1-p,b-0)+th(b,1)*pow(p,1)*pow(1-p,b-1));
    cout.precision(3);
    cout << r1 << endl;
    cout << r2;
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

double binomial(int x, int n, double p) {
    double q = 1.0 - p;
    int comb = 1;
    int end = n - x;
    int i;
    for (i=n; i>end; i--) {
        comb = comb * i;
    }
    for (i=x; i>0; i--) {
        comb = comb / i;
    }
    // printf("x=%d, n=%d, comb=%d, p^x=%f, q^(n-x)=%f ", x, n, comb, pow(p,x), pow(q,n-x));
    // printf("res=%f\n", comb * pow(p,x) * pow(q,n-x));
    return (comb * pow(p,x) * pow(q,n-x));
}

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    int reject_percent, batch_size;
    double p; // success probability
    double prejle2, prejge2;
    int i;
    
    scanf("%d %d", &reject_percent, &batch_size);
    p = 1 - (reject_percent/100.0);
    // printf("sucess prob p=%f\n", p);
    
    // no more than 2 rejects
    for (i = 0, prejle2 = 0.0; i <= 2; i++) {
        prejle2 += binomial(batch_size - i, batch_size, p);        
    }
    printf("%.3f\n", prejle2);
    
    // at least 2 rejects
    for (i = 2, prejge2 = 0.0; i <= batch_size; i++) {
        prejge2 += binomial(batch_size - i, batch_size, p);        
    }
    printf("%.3f\n", prejge2);
    
    return 0;
}

Problem solution in JavaScript programming.

function factorial (n) {
	var total = 1; 
	for (i=1; i<=n; i++) {
		total = total * i; 
	}
	return total; 
}

function a_sobre_b(a,b) {
    return factorial(a)/(factorial(b) * factorial(a-b));
}

function processData(input) {
    var suma_props = 0;
    for (var i=0; i<=2; i++) {
        suma_props += a_sobre_b(10,i)*Math.pow(0.12,i)*Math.pow(0.88,10-i);
    }
    console.log(suma_props.toFixed(3));
    suma_props = 0;
    for (var i=0; i<=1; i++) {
        suma_props += a_sobre_b(10,i)*Math.pow(0.12,i)*Math.pow(0.88,10-i);
    }
    console.log((1-suma_props).toFixed(3));
} 
 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});