Answer:
Step-by-step explanation:
Let's go ahead and list all the possible outcomes when a pair of dice is rolled;
![\begin{gathered} (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) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/sbg9r4zfc3p9o7z8xu0x2krzphmj6r4nfx.png)
We can see from the above that the total number of possible outcomes is 36
a) We're asked to determine the probability of rolling a sum not more than 3.
Let's list all the rolls that can produce a sum not more than 3;
![(1,1),(1,2),\text{and (2,1)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5jvzwhil9bdmdddc6lo98383u4z5dpvue0.png)
We can now find the probability as seen below;
![\begin{gathered} P(a\text{ sum not more than 3) =}\frac{Number\text{ of favourable outcomes}}{\text{Total number of possible outcomes}} \\ =(3)/(36) \\ =(1)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xiv9fggxj56xnrb8ymizzdyc5b96v8snbl.png)
b) To determine the probability of rolling a sum not less than 8, let's list all the rolls that can produce a sum not less than 8;
![\begin{gathered} (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3), \\ (6,4),(6,5),(6,6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fjb5cf6uybroh7vzskb1wlykl6aivs5css.png)
We can now find the probability as seen below;
![P(\text{not l}ess\text{ than 8)}=(15)/(36)=(5)/(12)](https://img.qammunity.org/2023/formulas/mathematics/high-school/ywckikl1u5hq7ou92i5e5onx1p5ibxtvqz.png)
c) To determine the probability of rolling a sum between 6 and 11(exclusive), let's list all the rolls that can produce a sum 7, 8, 9, and 10;
![\begin{gathered} (1,6),(2,5),(2,6),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,2),(5,3), \\ (5,4),(5,5),(6,1),(6,2),(6,3),(6,4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/f07q3g4f1sl07vi0jkp7so14zcoo69lixq.png)
We can now find the probability as seen below;
![P(a\text{ sum betwe}en\text{ 6 and 11(exclusive)})=(18)/(36)=(3)/(6)=(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/5dn5sdp2hd1wrkg927joedupu54q1fngn0.png)