Answer:
Probability is 21/36 = 7/12
Explanation:
Imagine we have two dice. Let's make one die red and one die green.
Each one of those dice can come up 1, 2, 3, 4, 5, or 6.
But only a certain number of rolls have the dice adding up to less than 8. We will call a result of 'less than 8' our 'desired outcome'
If the red die is 1, the green die can be 1, 2, 3, 4, 5, or 6. 6 results
If the red die is 2, the green die can be 1, 2, 3, 4, or 5. 5 results
If the red die is 3, the green die can be 1, 2, 3, or 4. 4 results
If the red die is 4, the green die can be 1, 2, or 3. 3 results
If the red die is 5, the green die can be 1 or 2. 2 results
If the red die is 6, the green die can be 1. 1 result
Total: 21 ways to get 'desired outcome'
To get the probability of a event, we divide the number of ways to get the desired outcome by the number of all possible outcomes.
So how many ways can the the dice come up?
If red is 1, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
If red is 2, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
If red is 3, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
If red is 4, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
If red is 5, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
If red is 6, green can be 1, 2, 3, 4, 5, or 6. - - > 6 ways
Total them up.
Or we could just multiply (6 ways for red) x (6 ways for green) = 36 ways
Total: 36 ways to get 'all possible outcomes'
So the probability of the desired outcome is 21/36, which simplifies to 7/12