Take a look at the table of possible outcomes when a pair of fair dice is rolled. what is the probability that the sum of the numbers rolled is either even or a multiple of 5?

Mark the entries in the table that are even. Those values are: 2, 4, 6, 8, 10, 12
There are 1+3+5+5+3+1 = 18 cases where this happens

Now mark the entries that are 5 or 10. There are multiples of 5. There are 4+3 = 7 cases of either 5 or 10. 

If we add up 18 and 7, we get 18+7 = 25. Now subtract out the cases where we get both an even number AND a multiple of 5. So the three tens are counted twice. We go from 25 to 25-3 = 22

Divide this value (22) over the total number of outcomes (6*6 = 36). Doing this gets us
22/36 = 11/18

The answer is choice A

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