Given an integer **N** denoting the number of dices, the task is to find the probability of the product of numbers appearing on the top faces of **N** thrown dices being a prime number. All **N** dices must be thrown simultaneously.

**Examples:**

Input: N = 2Output:6 / 36Explanation:

On throwing N(=2) dices simultaneously, the possible outcomes on the top faces of N(=2) dices having product equal to a prime number are: {(1, 2), (1, 3), (1, 5), (2, 1), (3, 1), (5, 1)}.

Therefore, the count of favourable outcomes = 6 and the count of the sample space is = 36

Therefore, the required output is (6 / 36)

Input:N = 3Output:9 / 216

**Naive Approach:** The simplest approach to solve this problem is to generate all possible outcomes on the top faces of **N** dices by throwing **N** dices simultaneously and for each possible outcome check if the product of numbers on the top faces is a prime number or not. If found to be true then increment the counter. Finally, print the probability of getting the product of numbers on the top faces as a prime number.

**Time Complexity:** O(6^{N} * N)**Auxiliary Space:** O(1)

**Efficient Approach:** To optimize the above approach the idea is to use the fact that the product of **N** number is a prime number only if **(N – 1)** numbers are **1** and a remaining number is a prime number. Following are the observations:

If the product of N numbers is a prime number then the value of (N – 1) numbers must be 1 and the remaining number must be a prime number.

Total count of prime numbers in the range [1, 6] is 3.

Therefore, the total number of outcomes in which the product of N numbers on the top faces as a prime number = 3 * N.

P(E) = N(E) / N(S)

P(E) = probability of getting the product of numbers on the top faces of N dices as a prime number.

N(E) = total count of favourable outcomes = 3 * N

N(S) = total number of events in the sample space = 6^{N}

Follow the steps below to solve this problem:

- Initialize a variable, say
**N_E**to store the count of favorable outcomes. - Initialize a variable, say
**N_S**to store the count of sample space. - Update
**N_E = 3 * N**. - Update
**N_S = 6**.^{N} - Finally, print the value of
**(N_E / N_S)**.

Below is the implementation of the above approach

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the value` `// of power(X, N)` `long` `long` `int` `power(` `long` `long` `int` `x,` ` ` `long` `long` `int` `N)` `{` ` ` `// Stores the value` ` ` `// of (X ^ N)` ` ` `long` `long` `int` `res = 1;` ` ` `// Calculate the value of` ` ` `// power(x, N)` ` ` `while` `(N > 0) {` ` ` ` ` `// If N is odd` ` ` `if` `(N & 1) {` ` ` ` ` `//Update res` ` ` `res = (res * x);` ` ` `}` ` ` ` ` `//Update x` ` ` `x = (x * x);` ` ` ` ` `//Update N` ` ` `N = N >> 1;` ` ` ` ` `}` ` ` `return` `res;` `}` `// Function to find the probability of` `// obtaining a prime number as the` `// product of N thrown dices` `void` `probablityPrimeprod(` `long` `long` `int` `N)` `{` ` ` `// Stores count of favorable outcomes` ` ` `long` `long` `int` `N_E = 3 * N;` ` ` ` ` `// Stores count of sample space` ` ` `long` `long` `int` `N_S = power(6, N);` ` ` ` ` `// Print the required probablity` ` ` `cout<<N_E<<` `" / "` `<<N_S;` ` ` `}` `// Driver code` `int` `main()` `{` ` ` `long` `long` `int` `N = 2;` ` ` `probablityPrimeprod(N);` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to find the value` `// of power(X, N)` `static` `int` `power(` `int` `x, ` `int` `N)` `{` ` ` ` ` `// Stores the value` ` ` `// of (X ^ N)` ` ` `int` `res = ` `1` `;` ` ` `// Calculate the value of` ` ` `// power(x, N)` ` ` `while` `(N > ` `0` `)` ` ` `{` ` ` ` ` `// If N is odd` ` ` `if` `(N % ` `2` `== ` `1` `)` ` ` `{` ` ` ` ` `// Update res` ` ` `res = (res * x);` ` ` `}` ` ` ` ` `// Update x` ` ` `x = (x * x);` ` ` ` ` `// Update N` ` ` `N = N >> ` `1` `;` ` ` `}` ` ` `return` `res;` `}` `// Function to find the probability of` `// obtaining a prime number as the` `// product of N thrown dices` `static` `void` `probablityPrimeprod(` `int` `N)` `{` ` ` ` ` `// Stores count of favorable outcomes` ` ` `int` `N_E = ` `3` `* N;` ` ` ` ` `// Stores count of sample space` ` ` `int` `N_S = power(` `6` `, N);` ` ` ` ` `// Print the required probablity` ` ` `System.out.print(N_E + ` `" / "` `+ N_S);` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `2` `;` ` ` ` ` `probablityPrimeprod(N);` `}` `}` `// This code is contributed by Amit Katiyar` |

## Python3

`# Python3 program to implement` `# the above approach` `# Function to find the value` `# of power(X, N)` `def` `power(x, N):` ` ` ` ` `# Stores the value` ` ` `# of (X ^ N)` ` ` `res ` `=` `1` ` ` `# Calculate the value of` ` ` `# power(x, N)` ` ` `while` `(N > ` `0` `):` ` ` `# If N is odd` ` ` `if` `(N ` `%` `2` `=` `=` `1` `):` ` ` ` ` `# Update res` ` ` `res ` `=` `(res ` `*` `x)` ` ` `# Update x` ` ` `x ` `=` `(x ` `*` `x)` ` ` `# Update N` ` ` `N ` `=` `N >> ` `1` ` ` `return` `res` `# Function to find the probability of` `# obtaining a prime number as the` `# product of N thrown dices` `def` `probablityPrimeprod(N):` ` ` ` ` `# Stores count of favorable outcomes` ` ` `N_E ` `=` `3` `*` `N` ` ` `# Stores count of sample space` ` ` `N_S ` `=` `power(` `6` `, N)` ` ` `# Prthe required probablity` ` ` `print` `(N_E, ` `" / "` `, N_S)` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `N ` `=` `2` ` ` `probablityPrimeprod(N)` `# This code is contributed by 29AjayKumar` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the` `// value of power(X, N)` `static` `int` `power(` `int` `x,` ` ` `int` `N)` `{ ` ` ` `// Stores the value` ` ` `// of (X ^ N)` ` ` `int` `res = 1;` ` ` `// Calculate the value` ` ` `// of power(x, N)` ` ` `while` `(N > 0)` ` ` `{` ` ` `// If N is odd` ` ` `if` `(N % 2 == 1)` ` ` `{` ` ` `// Update res` ` ` `res = (res * x);` ` ` `}` ` ` `// Update x` ` ` `x = (x * x);` ` ` `// Update N` ` ` `N = N >> 1;` ` ` `}` ` ` `return` `res;` `}` `// Function to find the probability` `// of obtaining a prime number as` `// the product of N thrown dices` `static` `void` `probablityPrimeprod(` `int` `N)` `{ ` ` ` `// Stores count of favorable` ` ` `// outcomes` ` ` `int` `N_E = 3 * N;` ` ` `// Stores count of sample` ` ` `// space` ` ` `int` `N_S = power(6, N);` ` ` `// Print the required` ` ` `// probablity` ` ` `Console.Write(N_E + ` `" / "` `+ N_S);` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 2;` ` ` `probablityPrimeprod(N);` `}` `}` `// This code is contributed by Princi Singh` |

## Javascript

`<script>` `// JavaScript program to implement the above approach` `// Function to find the value` `// of power(X, N)` `function` `power(x, N)` `{` ` ` ` ` `// Stores the value` ` ` `// of (X ^ N)` ` ` `let res = 1;` ` ` `// Calculate the value of` ` ` `// power(x, N)` ` ` `while` `(N > 0)` ` ` `{` ` ` ` ` `// If N is odd` ` ` `if` `(N % 2 == 1)` ` ` `{` ` ` ` ` `// Update res` ` ` `res = (res * x);` ` ` `}` ` ` ` ` `// Update x` ` ` `x = (x * x);` ` ` ` ` `// Update N` ` ` `N = N >> 1;` ` ` `}` ` ` `return` `res;` `}` `// Function to find the probability of` `// obtaining a prime number as the` `// product of N thrown dices` `function` `probablityPrimeprod(N)` `{` ` ` ` ` `// Stores count of favorable outcomes` ` ` `let N_E = 3 * N;` ` ` ` ` `// Stores count of sample space` ` ` `let N_S = power(6, N);` ` ` ` ` `// Print the required probablity` ` ` `document.write(N_E + ` `" / "` `+ N_S);` `}` `// Driver Code` ` ` `let N = 2;` ` ` `probablityPrimeprod(N);` ` ` ` ` `// This code is contributed by susmitakunndugoaldanga.` `</script>` |

**Output:**

6 / 36

**Time Complexity:** O(log_{2}N)**Auxiliary Space:** O( 1 )

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