Answer: 7/12
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Step-by-step explanation:
Let's list the ways to roll a 6.
- 1+5 = 6
- 2+4 = 6
- 3+3 = 6
- 4+2 = 6
- 5+1 = 6
There are A = 5 items listed above.
Now let's list the ways to roll a sum of 7.
- 1+6 = 7
- 2+5 = 7
- 3+4 = 7
- 4+3 = 7
- 5+2 = 7
- 6+1 = 7
There are B = 6 items listed above.
Lastly, let's list the ways to roll a sum of 8.
- 2+6 = 8
- 3+5 = 8
- 4+4 = 8
- 5+3 = 8
- 6+2 = 8
There are C = 5 items listed above.
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Summary so far:
We found there are
- A = 5 ways to roll a "6"
- B = 6 ways to roll a "7"
- C = 5 ways to roll an "8"
Therefore, we have A+B+C = 6+7+8 = 21 ways to roll either of those three sums. Each event is mutually exclusive.
This is out of 6*6 = 36 ways to roll two dice (whether we get those sums or not).
21/36 = (7*3)/(12*3) = 7/12 is the probability of getting any of those three sums mentioned.
7/12 = 0.5833 = 58.33% approximately