To find the probability of getting a sum of 3 or 8:
It is given that,
Two dice are rolled.
So, the total sample space, n(s)=36.
A be the event of getting a sum of 3.
So, A={(1,2), (2,1)}
n(A)=2
B be the event of getting a sum of 8.
So, B={(2,6), (3,5), (4,4), (5,3), (6,2)}
n(B)=5.
n(AnB)=0.
Using the formula,
![\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ =(n(A))/(n(S))+(n(A))/(n(S))-(n(A\cap B))/(n(S)) \\ =(2)/(36)+(5)/(36)-0 \\ =(7)/(36) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pamc92aohvpcowmypc773c8gk3s02hge8s.png)
Hence, the answer is,
![(7)/(36)](https://img.qammunity.org/2023/formulas/mathematics/college/durcjkzpmd0ww924orji5bzz3c2mjv9f8y.png)