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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
#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;
}Output
If run the above code it will generate the following output −
probability is : 0.1
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