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++
#include <bits/stdc++.h>
using namespace std;
float rainDayProbability( int a[], int n)
{
float count = 0, m;
for ( int i = 0; i < n; i++) {
if (a[i] == 1)
count++;
}
m = count / n;
return m;
}
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
import java.io.*;
import java.util.*;
class GFG
{
static float rainDayProbability( int a[],
int n)
{
float count = 0 , m;
for ( int i = 0 ; i < n; i++)
{
if (a[i] == 1 )
count++;
}
m = count / n;
return m;
}
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
def rainDayProbability(a, n) :
count = a.count( 1 )
m = count / n
return m
if __name__ = = "__main__" :
a = [ 1 , 0 , 1 , 0 , 1 , 1 , 1 , 1 ]
n = len (a)
print (rainDayProbability(a, n))
|
C#
using System;
class GFG
{
static float rainDayProbability( int []a,
int n)
{
float count = 0, m;
for ( int i = 0; i < n; i++)
{
if (a[i] == 1)
count++;
}
m = count / n;
return m;
}
public static void Main()
{
int []a = {1, 0, 1, 0,
1, 1, 1, 1};
int n = a.Length;
Console.WriteLine(rainDayProbability(a, n));
}
}
|
PHP
<?php
function rainDayProbability( $a , $n )
{
$count = 0; $m ;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $a [ $i ] == 1)
$count ++;
}
$m = $count / $n ;
return $m ;
}
$a = array (1, 0, 1, 0,
1, 1, 1, 1);
$n = count ( $a );
echo rainDayProbability( $a , $n );
?>
|
Javascript
<script>
function rainDayProbability(a,n)
{
let count = 0, m;
for (let i = 0; i < n; i++) {
if (a[i] == 1)
count++;
}
m = count / n;
return m;
}
let a = [1, 0, 1, 0, 1, 1, 1, 1 ];
let n = a.length;
document.write(rainDayProbability(a,n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
20 Aug, 2022
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