Solution
The table below is the required sample space of the to fair die
From the above table
The sample space contain 36 outcomes
Event A: The sum is greater than 9
we will look at the table and count all the elements that are greater than 9
There are 6 elements (they are 10, 10, 10, 11, 11, 12 from the table)
The probability for event A will be
![\begin{gathered} p(A)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(A)=\frac{6}{36} \\ p(A)=\frac{1}{6} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20p%28A%29%3D%5Cfrac%7B%5Ctext%7Brequired%20outcome%7D%7D%7B%5Ctext%7Btotal%20outcome%7D%7D%20%5C%5C%20p%28A%29%3D%5Cfrac%7B6%7D%7B36%7D%20%5C%5C%20p%28A%29%3D%5Cfrac%7B1%7D%7B6%7D%20%5Cend%7Bgathered%7D)
P(A) = 1/6
Event B: The sum is an even number.
We will look at the table and count the number of elements that are even
There are 18 elements (notice that there are 3 even number on each of the 6 rows of the table)
The probability for event B will be
![\begin{gathered} p(B)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(B)=\frac{18}{36} \\ p(B)=\frac{1}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20p%28B%29%3D%5Cfrac%7B%5Ctext%7Brequired%20outcome%7D%7D%7B%5Ctext%7Btotal%20outcome%7D%7D%20%5C%5C%20p%28B%29%3D%5Cfrac%7B18%7D%7B36%7D%20%5C%5C%20p%28B%29%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cend%7Bgathered%7D)
p(B) = 1/2