Probability of rain on N+1th day in C++

Given with an array containing 0’s and 1’s where 0’s represents no rain and 1’s represent rainy day. The task is to calculate the probability of rain on N+1th day.

To calculate the probability of rain on N+1th day we can apply the formula

Total number of rainy days in the set / total number of days in a

Input

arr[] = {1, 0, 0, 0, 1 }

Output

probability of rain on n+1th day : 0.4

Explanation

total number of rainy and non-rainy days are: 5
Total number of rainy days represented by 1 are: 2
Probability of rain on N+1th day is: 2 / 5 = 0.4

Input

arr[] = {0, 0, 1, 0}

Output

probability of rain on n+1th day : 0.25

Explanation

total number of rainy and non-rainy days are: 4
Total number of rainy days represented by 1 are: 1
Probability of rain on N+1th day is: 1 / 4 = 0.25

Approach used in the given program is as follows

Input the elements of an array
Input 1 for representing rainy day
Input 0 for representing non-rainy day
Calculate the probability by applying the formula given above
Print the result

Algorithm

Start
Step 1→ Declare Function to find probability of rain on n+1th day
   float probab_rain(int arr[], int size)
      declare float count = 0, a
      Loop For int i = 0 and i < size and i++
         IF (arr[i] == 1)
            Set count++
         End
      End
      Set a = count / size
      return a
step 2→ In main()
   Declare int arr[] = {1, 0, 0, 0, 1 }
   Declare int size = sizeof(arr) / sizeof(arr[0])
   Call probab_rain(arr, size)
Stop

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
//probability of rain on n+1th day
float probab_rain(int arr[], int size){
   float count = 0, a;
   for (int i = 0; i < size; i++){
      if (arr[i] == 1)
         count++;
      }
      a = count / size;
      return a;
}
int main(){
   int arr[] = {1, 0, 0, 0, 1 };
   int size = sizeof(arr) / sizeof(arr[0]);
   cout<<"probability of rain on n+1th day : "<<probab_rain(arr, size);
   return 0;
}