# Find probability of selecting element from kth column after N iterations

Given a matrix M of order N*M where Mij represent the occurrence of j in row i. The task is to find the probability of occurrence of given k in last row after applying the given operation:

• Starting from 1st row, we select one element from any column of the entire row, and add it to the same column of next row. We repeating this till the last row.

Examples:

```Input: k = 1, M[][] =
{{0, 1},
{1, 1}}

Output:0.666667

Input: k = 1, M[][] =
{{0, 1},
{1, 1}}

Output: 0.333333
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

• Pre-calculate a sum[] array which stores the sum of all elements of 1st row.
• Fill 1st row of dp[][] by the the first row of M[][] elements.
• Calculate the probability of selecting an element from each column of a particular row and add the same to the corresponding column.
• Also, update the Sum[] of the row while updating the value of dp[][] matrix.
• Finally, the value of dp[n][k] for given k is the required value for the probability of selecting element k after all iterations.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `#define n 4 ` `#define m 4 ` `using` `namespace` `std; ` ` `  `// Function to calculate probability ` `float` `calcProbability(``int` `M[][m], ``int` `k) ` `{ ` `    ``// declare dp[][] and sum[] ` `    ``float` `dp[m][n], sum[n]; ` ` `  `    ``// precalculate the first row ` `    ``for` `(``int` `j = 0; j < n; j++) { ` `        ``dp[0][j] = M[0][j]; ` `        ``sum[0] += dp[0][j]; ` `    ``} ` ` `  `    ``// calculate the probability for ` `    ``// each element and update dp table ` `    ``for` `(``int` `i = 1; i < m; i++) { ` `        ``for` `(``int` `j = 0; j < n; j++) { ` `            ``dp[i][j] += dp[i - 1][j] / sum[i - 1] + M[i][j]; ` `            ``sum[i] += dp[i][j]; ` `        ``} ` `    ``} ` ` `  `    ``// return result ` `    ``return` `dp[n - 1][k - 1] / sum[n - 1]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `M[m][n] = { { 1, 1, 0, 3 }, ` `                    ``{ 2, 3, 2, 3 }, ` `                    ``{ 9, 3, 0, 2 }, ` `                    ``{ 2, 3, 2, 2 } }; ` `    ``int` `k = 3; ` ` `  `    ``cout << calcProbability(M, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` ` `  `class` `GFG ` `{ ` `    ``final` `static` `int` `n = ``4` `;  ` `    ``final` `static` `int` `m = ``4` `; ` ` `  `    ``// Function to calculate probability  ` `    ``static` `float` `calcProbability(``int` `M[][], ``int` `k)  ` `    ``{  ` `        ``// declare dp[][] and sum[]  ` `        ``float` `dp[][] = ``new` `float``[m][n] ; ` `        ``float` `sum[] = ``new` `float``[n];  ` `     `  `        ``// precalculate the first row  ` `        ``for` `(``int` `j = ``0``; j < n; j++)  ` `        ``{  ` `            ``dp[``0``][j] = M[``0``][j];  ` `            ``sum[``0``] = sum[``0``] + dp[``0``][j];  ` `        ``}  ` `     `  `        ``// calculate the probability for  ` `        ``// each element and update dp table  ` `        ``for` `(``int` `i = ``1``; i < m; i++)  ` `        ``{  ` `            ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``{  ` `                ``dp[i][j] += dp[i - ``1``][j] / sum[i - ``1``] + ` `                            ``M[i][j];  ` `                ``sum[i] += dp[i][j];  ` `            ``}  ` `        ``}  ` `     `  `        ``// return result  ` `        ``return` `dp[n - ``1``][k - ``1``] / sum[n - ``1``];  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String []args) ` `    ``{  ` `        ``int` `M[][] = { { ``1``, ``1``, ``0``, ``3` `},  ` `                      ``{ ``2``, ``3``, ``2``, ``3` `},  ` `                      ``{ ``9``, ``3``, ``0``, ``2` `},  ` `                      ``{ ``2``, ``3``, ``2``, ``2` `} };  ` `        ``int` `k = ``3``;  ` `        ``System.out.println(calcProbability(M, k)); ` `         `  `    ``}  ` `} ` ` `  `// This code is contributed by Ryuga `

## Python3

 `# Python3 implementation of the  ` `# above approach ` ` `  `n ``=` `4` `m ``=` `4` ` `  `# Function to calculate probability ` `def` `calcProbability(M, k): ` ` `  `    ``# declare dp[][] and sum[] ` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n)]  ` `             ``for` `i ``in` `range``(m)] ` `    ``Sum` `=` `[``0` `for` `i ``in` `range``(n)] ` ` `  `    ``# precalculate the first row ` `    ``for` `j ``in` `range``(n):  ` `        ``dp[``0``][j] ``=` `M[``0``][j] ` `        ``Sum``[``0``] ``+``=` `dp[``0``][j] ` `     `  `    ``# calculate the probability for ` `    ``# each element and update dp table ` `    ``for` `i ``in` `range``(``1``, m):  ` `        ``for` `j ``in` `range``(n): ` `            ``dp[i][j] ``+``=` `(dp[i ``-` `1``][j] ``/`  `                         ``Sum``[i ``-` `1``] ``+` `M[i][j]) ` `            ``Sum``[i] ``+``=` `dp[i][j] ` `         `  `    ``# return result ` `    ``return` `dp[n ``-` `1``][k ``-` `1``] ``/` `Sum``[n ``-` `1``] ` ` `  `# Driver code ` ` `  `M ``=` `[[ ``1``, ``1``, ``0``, ``3` `], ` `     ``[ ``2``, ``3``, ``2``, ``3` `], ` `     ``[ ``9``, ``3``, ``0``, ``2` `], ` `     ``[ ``2``, ``3``, ``2``, ``2` `]]  ` `k ``=` `3` ` `  `print``(calcProbability(M, k)) ` ` `  `# This code is contributed ` `# by mohit kumar 29 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `n = 4 ;  ` `    ``static` `int` `m = 4 ; ` ` `  `    ``// Function to calculate probability  ` `    ``static` `float` `calcProbability(``int``[,] M, ``int` `k)  ` `    ``{  ` `        ``// declare dp[][] and sum[]  ` `        ``float``[,] dp = ``new` `float``[m,n] ; ` `        ``float``[] sum = ``new` `float``[n];  ` `     `  `        ``// precalculate the first row  ` `        ``for` `(``int` `j = 0; j < n; j++)  ` `        ``{  ` `            ``dp[0, j] = M[0, j];  ` `            ``sum[0] = sum[0] + dp[0, j];  ` `        ``}  ` `     `  `        ``// calculate the probability for  ` `        ``// each element and update dp table  ` `        ``for` `(``int` `i = 1; i < m; i++)  ` `        ``{  ` `            ``for` `(``int` `j = 0; j < n; j++) ` `            ``{  ` `                ``dp[i, j] += dp[i - 1,j] / sum[i - 1] + ` `                            ``M[i, j];  ` `                ``sum[i] += dp[i, j];  ` `            ``}  ` `        ``}  ` `     `  `        ``// return result  ` `        ``return` `dp[n - 1,k - 1] / sum[n - 1];  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``int``[,] M = { { 1, 1, 0, 3 },  ` `                    ``{ 2, 3, 2, 3 },  ` `                    ``{ 9, 3, 0, 2 },  ` `                    ``{ 2, 3, 2, 2 } };  ` `        ``int` `k = 3;  ` `        ``Console.Write(calcProbability(M, k)); ` `    ``}  ` `} ` ` `  `// This code is contributed by Ita_c. `

## PHP

Output:

```0.201212
```

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