A person starts walking from position X = 0, find the probability to reach exactly on X = N if she can only take either 2 steps or 3 steps. Probability for step length 2 is given i.e. P, probability for step length 3 is 1 – P.

**Examples :**

Input : N = 5, P = 0.20 Output : 0.32Explanation:- There are two ways to reach 5. 2+3 with probability = 0.2 * 0.8 = 0.16 3+2 with probability = 0.8 * 0.2 = 0.16 So, total probability = 0.32.

It is a simple dynamic programming problem. It is simple extension of this problem :- count-ofdifferent-ways-express-n-sum-1-3-4

Below is the implementation of the above approach.

## C++

// CPP Program to find probability to // reach N with P probability to take // 2 steps (1-P) to take 3 steps #include <bits/stdc++.h> using namespace std; // Returns probability to reach N float find_prob(int N, float P) { double dp[N + 1]; dp[0] = 1; dp[1] = 0; dp[2] = P; dp[3] = 1 - P; for (int i = 4; i <= N; ++i) dp[i] = (P)*dp[i - 2] + (1 - P) * dp[i - 3]; return dp[N]; } // Driver code int main() { int n = 5; float p = 0.2; cout << find_prob(n, p); return 0; }

## Java

// Java Program to find probability to // reach N with P probability to take // 2 steps (1-P) to take 3 steps import java.io.*; class GFG { // Returns probability to reach N static float find_prob(int N, float P) { double dp[] = new double[N + 1]; dp[0] = 1; dp[1] = 0; dp[2] = P; dp[3] = 1 - P; for (int i = 4; i <= N; ++i) dp[i] = (P) * dp[i - 2] + (1 - P) * dp[i - 3]; return ((float)(dp[N])); } // Driver code public static void main(String args[]) { int n = 5; float p = 0.2f; System.out.printf("%.2f",find_prob(n, p)); } } /* This code is contributed by Nikita Tiwari.*/

## Python3

# Python 3 Program to find # probability to reach N with # P probability to take 2 # steps (1-P) to take 3 steps # Returns probability to reach N def find_prob(N, P) : dp =[0] * (n + 1) dp[0] = 1 dp[1] = 0 dp[2] = P dp[3] = 1 - P for i in range(4, N + 1) : dp[i] = (P) * dp[i - 2] + (1 - P) * dp[i - 3] return dp[N] # Driver code n = 5 p = 0.2 print(round(find_prob(n, p), 2)) # This code is contributed by Nikita Tiwari.

## C#

// C# Program to find probability to // reach N with P probability to take // 2 steps (1-P) to take 3 steps using System; class GFG { // Returns probability to reach N static float find_prob(int N, float P) { double []dp = new double[N + 1]; dp[0] = 1; dp[1] = 0; dp[2] = P; dp[3] = 1 - P; for (int i = 4; i <= N; ++i) dp[i] = (P) * dp[i - 2] + (1 - P) * dp[i - 3]; return ((float)(dp[N])); } // Driver code public static void Main() { int n = 5; float p = 0.2f; Console.WriteLine(find_prob(n, p)); } } /* This code is contributed by vt_m.*/

## PHP

<?php // PHP Program to find probability to // reach N with P probability to take // 2 steps (1-P) to take 3 steps // Returns probability to reach N function find_prob($N, $P) { $dp; $dp[0] = 1; $dp[1] = 0; $dp[2] = $P; $dp[3] = 1 - $P; for ($i = 4; $i <= $N; ++$i) $dp[$i] = ($P) * $dp[$i - 2] + (1 - $P) * $dp[$i - 3]; return $dp[$N]; } // Driver code $n = 5; $p = 0.2; echo find_prob($n, $p); // This code is contributed by mits. ?>

**Output :**

0.32

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.