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

https://www.geeksforgeeks.org/probability-reaching-point-2-3-steps-time/

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 : 0.32
Explanation :-
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

    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]));

    }