Probability of reaching a point with 2 or 3 steps at a time in C++

A person “A” is walking from a starting position X = 0, the task is to find the probability to reach exactly X = num, if he/she can either take 2 or 3 steps. Probability for step length 2 i.e. P, the probability for the step length 3 is 1 - P.

Input

num = 5, p = 0.2

Output

0.32

Explanation

There can be 2 ways to reach num, i.e, 5
2+3 with probability 0.2 * 0.8 = 0.16
3+2 with probability 0.8 * 0.2 = 0.16
So, total probability will be 0.16 + 0.16 = 0.32

Input

num = 2, p = 0.1

Output

0.1

Approach used below is as follows to solve the problem

We will be using dynamic programming approach to solve the problem.

In the solution we will −

Declare a probability array whose size is num+1 and assign its values as, Set probab[0] = 1, Set probab[1] = 0, Set probab[2] = p, Set probab[3] = 1 – p
Iterate i from 0 to num, while increment its value
For every i, Set probab[i] = (p)*probab[i - 2] + (1 - p) * probab[i - 3]
Return the probab[num]
Print result.

Algorithm

Start
Step 1→ declare function to calculate probability of reaching a point with 2 or 3 steps
at a time
   float probab(int num, float p)
      Declare double probab[num + 1]
         `Set probab[0] = 1
         Set probab[1] = 0
         Set probab[2] = p
         Set probab[3] = 1 – p
         Loop For int i = 4 and i <= num and ++i
            Set probab[i] = (p)*probab[i - 2] + (1 - p) * probab[i - 3]
         End
         return probab[num]
Step 2→ In main()
   Declare int num = 2
   Declare float p = 0.1
   Call probab(num, p)
Stop

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
//function to calculate probability of reaching a point with 2 or 3 steps at a time
float probab(int num, float p){
   double probab[num + 1];
   probab[0] = 1;
   probab[1] = 0;
   probab[2] = p;
   probab[3] = 1 - p;
   for (int i = 4; i <= num; ++i)
      probab[i] = (p)*probab[i - 2] + (1 - p) * probab[i - 3];
   return probab[num];
}
int main(){
   int num = 2;
   float p = 0.1;
   cout<<"probability is : "<<probab(num, p);
   return 0;
}