Thread: Dice: chance and odds

I am trying to help my middle schooler with this problem:

If a pair of dice is tossed once:

(a) What is the chance of rolling either a 4 or 7?

(b) What are the odds of rolling a total of 12?

Write down all possible ways of rolling each number (keep track of the results of the die rolls individually, so the left die showing a 1 and the right die showing a 3 is different from the left die showing a 3 and the right die showing a 1):
2: (1,1)
3: (1,2) (2,1)
4: (1,3) (2,2) (3,1)
5: (1,4) (2,3) (3,2) (4,1)
6: (1,5) (2,4) (3,3) (4,2) (5,1)
7: (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
8: (2,6) (3,5) (4,4) (5,3) (6,2)
9: (3,6) (4,5) (5,4) (6,3)
10: (4,6) (5,5) (6,4)
11: (5,6) (6,5)
12: (6,6)

This gives you all possible outcomes. There are 36 possible outcomes, each equally likely. Among those 36, there are three outcomes that yield a total of 4 among the two dice and six outcomes that yield 7 among the two dice. Hence, there are three plus six equals nine outcomes out of all 36 possible outcomes where the result is either a 4 or a 7. That means the chance of rolling either a 4 or 7 is $\displaystyle \dfrac{9}{36} = \dfrac{1}{4} = 0.25 = 25\%$

There is only one possible outcome that results in a 12 being rolled. So, the chance of rolling a 12 is $\displaystyle \dfrac{1}{36} = 0.02777... \approx 2.78\%$