Given:
Two dice are rolled together.
Total number of possible outcomes.
To find:
The list of total possible outcomes.
The probability of getting a sum of 11 in these outcomes.
The probability of getting a sum less than or equal to 4.
The probability of getting a sum of 13 or more.
Solution:
If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is
S = {(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)}
Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3
The probability of getting a sum of 11 in these outcomes is
![P(\text{sum=11})=\dfrac{3}{36}](https://tex.z-dn.net/?f=P%28%5Ctext%7Bsum%3D11%7D%29%3D%5Cdfrac%7B3%7D%7B36%7D)
![P(\text{sum=11})=\dfrac{1}{12}](https://tex.z-dn.net/?f=P%28%5Ctext%7Bsum%3D11%7D%29%3D%5Cdfrac%7B1%7D%7B12%7D)
Therefore, the probability of getting a sum of 11 in these outcomes is
.
Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6
The probability of getting a sum less than or equal to 4 is
![P({sum\leq 4})=\dfrac{6}{36}](https://tex.z-dn.net/?f=P%28%7Bsum%5Cleq%204%7D%29%3D%5Cdfrac%7B6%7D%7B36%7D)
![P({sum\leq 4})=\dfrac{1}{6}](https://tex.z-dn.net/?f=P%28%7Bsum%5Cleq%204%7D%29%3D%5Cdfrac%7B1%7D%7B6%7D)
Therefore, the probability of getting a sum less than or equal to 4 is
.
Sum of 13 or more = empty set because maximum sum is 12.
The probability of getting a sum of 13 or more is
![P(sum\geq 13)=\dfrac{0}{36}](https://tex.z-dn.net/?f=P%28sum%5Cgeq%2013%29%3D%5Cdfrac%7B0%7D%7B36%7D)
![P({sum\geq 13})=0](https://tex.z-dn.net/?f=P%28%7Bsum%5Cgeq%2013%7D%29%3D0)
Therefore, the probability of getting a sum of 13 or more is 0.