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user2974951
user2974951
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For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers (assuming they can be any number from 1 to 6, which your problem seems to state that they cannot). Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.

For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers. Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.

For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers (assuming they can be any number from 1 to 6, which your problem seems to state that they cannot). Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.

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user2974951
user2974951
  • 8k
  • 2
  • 20
  • 32

For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers. Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.