WILL MARK BRAINLIST!! Olivia rolls two fair number cubes numbered from 1 to 6. She first defines the sample space as shown below: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Based on the sample space, what is the probability of getting a total of 10?

The probability of getting a total of 10 will be [tex] \frac{1}{12} [/tex]ExplanationThe total possible sample spaces for rolling two fair numbered cubes are : (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)There are total 36 sample spaces. We need to get a sum of 10 , so the possible sample spaces will be: (4, 6) , (5, 5) and (6, 4)That means there are 3 possible sample spaces for getting a sum of 10. So, the probability of getting a total of 10 [tex] = \frac{3}{36} = \frac{1}{12} [/tex]