# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ code to find the probability of rain ` `// on n+1-th day when previous day's data is given ` `#include ` `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; ` `} `

## Java

 `// 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)); ` `} ` `} `

## Python 3

 `# 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 `

## C#

 `// 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. `

## PHP

 ` `

Output:

```0.75
```

Time Complexity: O(N)

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Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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Improved By : inderDuMCA, AnkitRai01