掷出 2 个骰子 N 次得到和的概率
考虑到总数。任务是找出两个骰子掷 N 次的和出现的概率。 概率被定义为对结果总数有利的结果数。概率总是介于 0 和 1 之间。 示例:
Input: sum = 11, times = 1
Output: 2 / 36
favorable outcomes = (5, 6) and (6, 5) i.e 2
Total outcomes = (1, 1), (1, 2), (1, 3)...(6, 6) i.e 36
Probability = (2 / 36)
Input: sum = 7, times = 7
Output: 1 / 279936
公式:-
掷出 2 个骰子 n 次出现和的概率= (有利/总)^ N
进场:-
首先，计算 1 次掷出 2 个骰子的总和出现的概率。 假设概率 1。 现在，计算两个骰子掷 n 次之和出现的概率为: 概率 2 =(概率 1) ^ N. 即概率 1 升 n 次方
以下是上述方法的实施:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// function that calculates Probability.
int Probability(int sum, int times)
{
float favorable = 0.0, total = 36.0;
long int probability = 0;
// To calculate favorable outcomes
// in thrown of 2 dices 1 times.
for (int i = 1; i <= 6; i++) {
for (int j = 1; j <= 6; j++) {
if ((i + j) == sum)
favorable++;
}
}
int gcd1 = __gcd((int)favorable, (int)total);
// Reduce to simplest Form.
favorable = favorable / (float)gcd1;
total = total / (float)gcd1;
// Probability of occurring sum on 2 dice N times.
probability = pow(total, times);
return probability;
}
// Driver Code
int main()
{
int sum = 7, times = 7;
cout << "1"
<< "/" << Probability(sum, times);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.io.*;
class GFG
{
// Recursive function to return
// gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// function that calculates
// Probability.
static long Probability(int sum,
int times)
{
float favorable = 0, total = 36;
long probability = 0;
// To calculate favorable outcomes
// in thrown of 2 dices 1 times.
for (int i = 1; i <= 6; i++)
{
for (int j = 1; j <= 6; j++)
{
if ((i + j) == sum)
favorable++;
}
}
int gcd1 = __gcd((int)favorable,
(int)total);
// Reduce to simplest Form.
favorable = favorable / (float)gcd1;
total = total / (float)gcd1;
// Probability of occurring
// sum on 2 dice N times.
probability = (long)Math.pow(total, times);
return probability;
}
// Driver Code
public static void main (String[] args)
{
int sum = 7, times = 7;
System.out.println( "1" + "/" +
Probability(sum, times));
}
}
// This code is contributed
// by inder_verma
Python 3
# Python 3 implementation of above approach
# from math import everything
from math import *
# function that calculates Probability.
def Probability(sum, times) :
favorable, total, probability = 0.0, 36.0, 0
# To calculate favorable outcomes
# in thrown of 2 dices 1 times.
for i in range(7) :
for j in range(7) :
if ((i + j) == sum) :
favorable += 1
gcd1 = gcd(int(favorable), int(total))
# Reduce to simplest Form.
favorable = favorable / gcd1
total = total / gcd1
# Probability of occurring sum on 2 dice N times.
probability = pow(total, times)
return int(probability)
# Driver Code
if __name__ == "__main__" :
sum, times = 7, 7
print("1","/",Probability(sum, times))
# This code is contributed by ANKITRAI1
C
// C# implementation of above approach
class GFG
{
// Recursive function to return
// gcd of a and b
static int __gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// function that calculates
// Probability.
static long Probability(int sum,
int times)
{
float favorable = 0, total = 36;
long probability = 0;
// To calculate favorable outcomes
// in thrown of 2 dices 1 times.
for (int i = 1; i <= 6; i++)
{
for (int j = 1; j <= 6; j++)
{
if ((i + j) == sum)
favorable++;
}
}
int gcd1 = __gcd((int)favorable,
(int)total);
// Reduce to simplest Form.
favorable = favorable / (float)gcd1;
total = total / (float)gcd1;
// Probability of occurring
// sum on 2 dice N times.
probability = (long)System.Math.Pow(total, times);
return probability;
}
// Driver Code
public static void Main()
{
int sum = 7, times = 7;
System.Console.WriteLine( "1" + "/" +
Probability(sum, times));
}
}
// This code is contributed
// by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// function to calculate gcd
function getGCDBetween($a, $b)
{
while ($b != 0)
{
$m = $a % $b;
$a = $b;
$b = $m;
}
return $a;
}
// function that calculates Probability.
function Probability($sum, $times)
{
$favorable = 0.0;
$total = 36.0;
$probability = 0;
// To calculate favorable outcomes
// in thrown of 2 dices 1 times.
for ($i = 1; $i <= 6; $i++)
{
for ($j = 1; $j <= 6; $j++)
{
if (($i + $j) == $sum)
$favorable++;
}
}
$gcd1 = getGCDBetween((int)$favorable,
(int)$total);
// Reduce to simplest Form.
$favorable = $favorable / (float)$gcd1;
$total = $total / (float)$gcd1;
// Probability of occurring
// sum on 2 dice N times.
$probability = pow($total, $times);
return $probability;
}
// Driver Code
$sum = 7;
$times = 7;
echo "1", "/", Probability($sum, $times);
// This code is contributed by ash264
?>
java 描述语言
<script>
// Javascript implementation of above approach
// Recursive function to return
// gcd of a and b
function __gcd(a, b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// function that calculates Probability.
function Probability(sum, times)
{
var favorable = 0.0, total = 36.0;
var probability = 0;
// To calculate favorable outcomes
// in thrown of 2 dices 1 times.
for (var i = 1; i <= 6; i++) {
for (var j = 1; j <= 6; j++) {
if ((i + j) == sum)
favorable++;
}
}
var gcd1 = __gcd(favorable, total);
// Reduce to simplest Form.
favorable = favorable / gcd1;
total = total / gcd1;
// Probability of occurring sum on 2 dice N times.
probability = Math.pow(total, times);
return probability;
}
// Driver Code
var sum = 7, times = 7;
document.write( "1"
+ "/" + Probability(sum, times));
</script>
Output:
1/279936
版权属于：月萌API www.moonapi.com，转载请注明出处
