If three dice are tossed, their sum could be 7. What is the probability of this?

I'm guessing it's something to do with combinations?? I wrote down all the combinations of it equaling 7:
3, 3, 1 // 4, 2, 1 // 1, 5, 1
So I thought that since there are five possible numbers the three dice could be, it would be 5choose1 times 4choose1 times 3choose1, but I don't think that's right...

You left out the case 2,2,3

I would go with permutations, the total number of ways for the 3 dice to fall
= 6*6*6 = 216
To get a 7:
permute 223 in 3!/2! or 3 ways
permute 331 in 3 ways
permute 115 in 3 ways
permute 124 in 6 ways, for a total of 15 ways to get a 7

so prob(7) = 15/216
= 5/72

To find the probability of the sum of three dice being 7, you can use the concept of combinations.

First, let's analyze the possible combinations that result in a sum of 7:

1. (1, 1, 5)
2. (1, 2, 4)
3. (1, 3, 3)
4. (1, 4, 2)
5. (1, 5, 1)
6. (2, 1, 4)
7. (2, 2, 3)
8. (2, 3, 2)
9. (2, 4, 1)
10. (3, 1, 3)
11. (3, 2, 2)
12. (3, 3, 1)
13. (4, 1, 2)
14. (4, 2, 1)
15. (5, 1, 1)

There are 15 different combinations that result in a sum of 7 when tossing three dice.

Now, let's calculate the probability. The total number of possible outcomes when tossing three dice is 6^3 = 216 (since each die has 6 possible outcomes). Therefore, the probability is:

P(sum of 7) = Number of favorable outcomes / Total number of outcomes

P(sum of 7) = 15 / 216

Simplifying this fraction gives the probability:

P(sum of 7) = 5 / 72

So, the probability of obtaining a sum of 7 when tossing three dice is 5/72.

To find the probability of getting a sum of 7 when three dice are tossed, we can use a combination of counting the favorable outcomes and the total number of possible outcomes.

First, let's calculate the total number of possible outcomes when three dice are tossed. Since each die has 6 sides, the total number of outcomes for one die is 6. Therefore, the total number of outcomes for three dice is 6 * 6 * 6 = 216.

To find the favorable outcomes, let's consider the combinations that can give us a sum of 7. We can break it down into the following scenarios:

1. 1, 1, 5
2. 1, 2, 4
3. 1, 3, 3
4. 2, 1, 4
5. 2, 2, 3
6. 2, 3, 2
7. 3, 1, 3
8. 3, 2, 2
9. 4, 1, 2
10. 4, 2, 1
11. 5, 1, 1

Now that we have identified the favorable outcomes, we can count them. There are 11 favorable outcomes.

Therefore, the probability of getting a sum of 7 when three dice are tossed is:

Probability = (Number of favorable outcomes) / (Total number of outcomes)
Probability = 11 / 216

Simplifying this fraction further is optional.