We have 4 Sports magazines, and 4 Fashion magazines
We have to select 3 magazines
Total ways of selecting 3 magazines out of these 8 -> 8C3 (8 choose 3) = 8!/(3! * 5!) = 56
Now we need probability that "at least one" of the fashion magazines will be selected. This scenario would include: selection of exactly 1 fashion magazine, exactly 2 fashion magazines, or exactly 3 fashion magazines. Either we can calculate these individual probabilities and add them up, or we can find out the probability that no fashion magazine is selected, and subtract it from 1, since
1 = Probability(select 0 fashion magazines) + probability(select 1 fashion magazine) + probability(select 2 fashion magazine) + probability(Select 3 fashion magazine)

Now P(select no fashion magazine) = ways of selecting only sports magazines / total ways = 4C3 / 56
= 4/56 = 1/14

So our desired probability = 1 - 1/14 = 13/14