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C++ Program Probability for three randomly chosen numbers to be in AP
Given with an array of numbers ‘n’ and the task is to find the probability of three randomly chosen numbers to be in AP.
Example
Input-: arr[] = { 2,3,4,7,1,2,3 }
Output-: Probability of three random numbers being in A.P is: 0.107692
Input-:arr[] = { 1, 2, 3, 4, 5 }
Output-: Probability of three random numbers being in A.P is: 0.151515Approach used in the below program is as follows −
- Input the array of positive integers
- Calculate the size of the array
Apply the formula given below to find the probability of three random numbers to be in AP
3 n / (4 (n * n) – 1)
- Print the result
ALGORITHM
Start
Step 1-> function to calculate the probability of three random numbers be in AP
double probab(int n)
return (3.0 * n) / (4.0 * (n * n) - 1)
Step 2->In main()
declare an array of elements as int arr[] = { 2,3,4,7,1,2,3 }
calculate size of an array as int size = sizeof(arr)/sizeof(arr[0])
call the function to calculate probability as probab(size)
StopExample
#include <bits/stdc++.h>
using namespace std;
//calculate probability of three random numbers be in AP
double probab(int n) {
return (3.0 * n) / (4.0 * (n * n) - 1);
}
int main() {
int arr[] = { 2,3,4,7,1,2,3 };
int size = sizeof(arr)/sizeof(arr[0]);
cout<<"probability of three random numbers being in A.P is : "<<probab(size);
return 0;
}Output
Probability of three random numbers being in A.P is: 0.107692
Updated on: 20-Nov-2019
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