Question

using basic c++

3. A certain 8-sided die is weighted such that 7's and 8's come up half as likely and 2's and 4's come up twice as often as any other number. Write a program to simulate 500 throws of this [8] die and output the frequency with which each number occurs. Total marks: 8

Answer 1

#include <iostream> 1 #include <time.h> 2 using namespace std 4 int main) srand(time (NULL)); int dice[1000] 6 {0, е, е, е, в, е, е, е}; 7 for (int i-0; i<500; it+) 10 int one = rand) 18; = 0 | 11 if (one dice[0] else if (one dice [4]+ else if (one dice[1] else if (one dice[2]+ else if (one dice [3]+ else if (one dice[5] else if (one dice[6] else if(one dice [7] 1) 12 one == 13 == 2 | 3) 14 one 15 =6 || one 7) == 4| | 16 one 5 one 17 8| 9) 18 one = == 19 11 || one 12 || one == 13) 10 | | 20 one = 21 14 15) 22 one= 23 16) 24 == 25 17) 26 27 28 for (int i-0; i<8; i++) cout < i+1 << 29 dice[i times." << endl; 30 occurred << 31

#include <iostream>

#include <time.h>

using namespace std;

int main() {

srand(time(NULL));

int dice[1000] = {0, 0, 0, 0, 0, 0, 0, 0};

for(int i=0; i<500; i++)

{

int one = rand() % 18;

if(one == 0 || one == 1)

dice[0]++;

else if(one == 2 || one == 3)

dice[4]++;

else if(one == 4 || one == 5 || one == 6 || one == 7)

dice[1]++;

else if(one == 8 || one == 9)

dice[2]++;

else if(one == 10 || one == 11 || one == 12 || one == 13)

dice[3]++;

else if(one == 14 || one == 15)

dice[5]++;

else if(one == 16)

dice[6]++;

else if(one == 17)

dice[7]++;

}

for(int i=0; i<8; i++)

cout << i+1 << " occurred " << dice[i] << " times." << endl;

}

/*OUTPUT

1 occurred 60 times.

2 occurred 99 times.

3 occurred 55 times.

4 occurred 109 times.

5 occurred 55 times.

6 occurred 58 times.

7 occurred 30 times.

8 occurred 34 times.

*/

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