Probability of rain on N+1th day

Given an array of 1 and 0’s, where Ai = 1 denotes that ith day was a rainy day and Ai = 0 denotes it was not a rainy day. The task is to find the probability that the N+1th was a rainy day.

Examples:

Input: a[] = {0, 0, 1, 0}
Output: .25
Since one day was rainy out of 4 days, hence the probability on
5th day will be 0.25

Input: a[] = {1, 0, 1, 0, 1, 1, 1}
Output: 0.71

The probability of rain on N+1th day can be found out using the below formula:

Probability = number of rainy days / total number of days.

First, count the number of 1’s and then the probability will be the number of 1’s divided by N i.e. count / N.

Below is the implementation of the above approach:

C++

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// C++ code to find the probability of rain
// on n+1-th day when previous day's data is given
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the probability
float rainDayProbability(int a[], int n)
{
    float count = 0, m;
  
    // count 1
    for (int i = 0; i < n; i++) {
        if (a[i] == 1)
            count++;
    }
  
    // find probability
    m = count / n;
    return m;
}
  
// Driver Code
int main()
{
  
    int a[] = { 1, 0, 1, 0, 1, 1, 1, 1 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << rainDayProbability(a, n);
    return 0;
}

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Java

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// Java code to find the
// probability of rain
// on n+1-th day when previous 
// day's data is given
import java.io.*;
import java.util.*;
  
class GFG
{
      
// Function to find 
// the probability
static float rainDayProbability(int a[],
                                int n)
{
    float count = 0, m;
  
    // count 1
    for (int i = 0; i < n; i++) 
    {
        if (a[i] == 1)
            count++;
    }
  
    // find probability
    m = count / n;
    return m;
}
  
// Driver Code
public static void main(String args[])
{
    int a[] = { 1, 0, 1, 0, 1, 1, 1, 1 };
    int n = a.length;
  
    System.out.print(rainDayProbability(a, n));
}
}

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Python 3

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# Python 3 program to find
# the probability of rain 
# on n+1-th day when previous 
# day's data is given 
  
# Function to find the probability 
def rainDayProbability(a, n) :
  
    # count occurence of 1
    count = a.count(1)
  
    # find probability
    m = count / n
      
    return m
  
# Driver code
if __name__ == "__main__" :
  
    a = [ 1, 0, 1, 0, 1, 1, 1, 1]
    n = len(a)
  
    # function calling
    print(rainDayProbability(a, n))
  
# This code is contributed
# by ANKITRAI1

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C#

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// C# code to find the
// probability of rain
// on n+1-th day when 
// previous day's data
// is given
using System;
  
class GFG
{
      
// Function to find 
// the probability
static float rainDayProbability(int []a,
                                int n)
{
    float count = 0, m;
  
    // count 1
    for (int i = 0; i < n; i++) 
    {
        if (a[i] == 1)
            count++;
    }
  
    // find probability
    m = count / n;
    return m;
}
  
// Driver Code
public static void Main()
{
    int []a = {1, 0, 1, 0, 
               1, 1, 1, 1};
    int n = a.Length;
  
    Console.WriteLine(rainDayProbability(a, n));
}
}
  
// This code is contributed 
// by inder_verma.

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PHP

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<?php
// PHP code to find the 
// probability of rain
// on n+1-th day when 
// previous day's data 
// is given
  
// Function to find
// the probability
function rainDayProbability($a, $n)
{
    $count = 0; $m;
  
    // count 1
    for ($i = 0; $i <$n; $i++) 
    {
        if ($a[$i] == 1)
            $count++;
    }
  
    // find probability
    $m = $count / $n;
    return $m;
}
  
// Driver Code
$a = array(1, 0, 1, 0, 
           1, 1, 1, 1);
$n = count($a);
  
echo rainDayProbability($a, $n);
  
// This code is contributed 
// by inder_verma.
?>

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Output:

0.75

Time Complexity: O(N)



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Improved By : inderDuMCA, AnkitRai01