原文:
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
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）
java 描述语言
output:
1/279936
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