Problem A
The quicker way of doing this is to figure out how many ways you can get 5 or under. That is much easier than 6 and above. The first thing you must do is calculate how many ways are possible to throw the same die twice.
Step One
What is the total number of ways of throwing a die twice.
The first time you throw the die, you can get 1 of 6 numbers. So the first time you throw, the results of what you do get is 1/6.
The second time you throw it's the same thing.
So what you get is 1/6 * 1/6 or 1/36 times that you can get a specified result. Put another way, there are 36 ways that a dice can be thrown twice.
Step Two
How many ways will you throw less than 6?
2: 1 and 1 or 2. There is only 1 way you can throw a two.
3: You can throw a 3 in two different ways. The first time you can throw a 1 the second time a 2 or the other way around.
4: You can throw that 3 different a four 2 and 2 or 1 and 3 or 3 and 1.
5: You can throw that 4 different ways. (3 and 2 or 2 and 3) or (4 and 1 or 1 and 4)
The total number of ways for under 6 is
4 + 3 +2 + 1 = 10
So there are 10 / 36 ways of getting under 6.
Step Three
How many ways are there of throwing 6 and over?
if there are 10 ways of throwing under 6, there must be 36 – 10 = 26 ways of throwing 6 and over
The answer to problem A is 26/36 = 13 / 18. <<<<<< Answer.
Problem B
Again, the easy way is to figure out the number of ways that you can throw something that is divisible 3 or 5.
Divisible by three
3 6 9 12 are all divisible by 3
3: there are 2 ways to throw a 3 from the question above.
6: there are 5 ways to throw a 6 (1 and 5 and 5 and 1) or (4 and 2 or 2 and 4) or (3 and 3)
9 there are 4 ways to throw a 9 (4 and 5 or 5 and 4) or (6 and 3) and (3 and 6)
12 there is only 1 way to throw a 12 (6 and 6)
For divisibility by 5 there are only 2 numbers 5 and 10
5 can be thown 4 different ways. (see above)
10: can be thown 3 different ways. (see if you can figure out how).
Total 2 + 5 + 4 + 1 + 4 + 3 = 19 So there are 19 ways of throwing something divisible by 3 or 5.
Therefore there are 36 – 19 = 17 other numbers can can be thown and the answer for B is
17/36 <<<<< answer
Problem A
The quicker way of doing this is to figure out how many ways you can get 5 or under. That is much easier than 6 and above. The first thing you must do is calculate how many ways are possible to throw the same die twice.
Step One
What is the total number of ways of throwing a die twice.
The first time you throw the die, you can get 1 of 6 numbers. So the first time you throw, the results of what you do get is 1/6.
The second time you throw it's the same thing.
So what you get is 1/6 * 1/6 or 1/36 times that you can get a specified result. Put another way, there are 36 ways that a dice can be thrown twice.
Step Two
How many ways will you throw less than 6?
2: 1 and 1 or 2. There is only 1 way you can throw a two.
3: You can throw a 3 in two different ways. The first time you can throw a 1 the second time a 2 or the other way around.
4: You can throw that 3 different a four 2 and 2 or 1 and 3 or 3 and 1.
5: You can throw that 4 different ways. (3 and 2 or 2 and 3) or (4 and 1 or 1 and 4)
The total number of ways for under 6 is
4 + 3 +2 + 1 = 10
So there are 10 / 36 ways of getting under 6.
Step Three
How many ways are there of throwing 6 and over?
if there are 10 ways of throwing under 6, there must be 36 – 10 = 26 ways of throwing 6 and over
The answer to problem A is 26/36 = 13 / 18. <<<<<< Answer.
Problem B
Again, the easy way is to figure out the number of ways that you can throw something that is divisible 3 or 5.
Divisible by three
3 6 9 12 are all divisible by 3
3: there are 2 ways to throw a 3 from the question above.
6: there are 5 ways to throw a 6 (1 and 5 and 5 and 1) or (4 and 2 or 2 and 4) or (3 and 3)
9 there are 4 ways to throw a 9 (4 and 5 or 5 and 4) or (6 and 3) and (3 and 6)
12 there is only 1 way to throw a 12 (6 and 6)
For divisibility by 5 there are only 2 numbers 5 and 10
5 can be thown 4 different ways. (see above)
10: can be thown 3 different ways. (see if you can figure out how).
Total 2 + 5 + 4 + 1 + 4 + 3 = 19 So there are 19 ways of throwing something divisible by 3 or 5.
Therefore there are 36 – 19 = 17 other numbers can can be thown and the answer for B is
17/36 <<<<< answer
