The question says that you roll the pair of dice three times, and you want the probability that you get a sum of 5 at least once.
Let X = the sum of the 2 dice.
P(X=5) = 4/36 = 1/9 (4 ways to get 5, 36=62 different results of 2 dice )
P(X≠5) = 1 - 1/9 = 8/9
P(at least one 5) = 1 - P(No 5's) (see the latest video I posted here for explanation of why this is so)
For three rolls:
P(at least one 5) = 1 - [P(X≠5)]3 = 1 - (8/9)3 ≈ .2977