Consider 2 dice (fair). What is the probability that a 7 or an 11 will turn out?
How many ways can you roll a sum of 7 or 11
(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)
Notice the diagonals
Lets do the 7's first
(6,1) (5,2) (4,3) (3,4) (2,5) (1,6)
Now the 11's (6,5) and (5,6)
so the are eight total ways and thirty six possible outcome so
$\displaystyle P(7 \cup 11)=\frac{8}{36}=\frac{2}{9}$
Each die has 6 sides: numbers 1 through 6. One set of outcomes for the pair of dice are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6). You will have similar sets if you replace the 1s in the first place with the numbers 2 through 6. So you will have 5 other similar sets, which will give you a total of 6 sets. So you have 6 sets with 6 particular outcomes. This give you a total of 36 possible outcomes. First part done.
Now you need to determine the number of times you toss the pair of dice such that their sum is either 7 or 11. (Remember for "or" you add the probabilities.) So for the sum of 7 you have (1,6), (2,5), (3,4). Note however that you have 3 *2=6 possibilities for 7, i.e. you also have the possibilities (6,1), (5,2), (3,4) (order matters=permutation). You can solve for the sum of 11.
Now you can determine the probability by dividing the sum of possibilities of 7 or 11 by the total number of possibilities.
Let me know if this is unclear to you.
Happy studying!
-Yvonne