To find the number of different selections of 5 marbles containing at least 4 marbles of the same color, we consider two cases:

1. 4 red marbles and 1 blue marble:

   The number of ways to choose 4 red marbles from 8 is given by \(\binom{8}{4}\), and the number of ways to choose 1 blue marble from 4 is \(\binom{4}{1}\).

   Thus, the total number of ways for this case is \(\binom{8}{4} \times \binom{4}{1} = 70 \times 4 = 280\).

2. 4 blue marbles and 1 red marble:

   The number of ways to choose 4 blue marbles from 4 is \(\binom{4}{4}\), and the number of ways to choose 1 red marble from 8 is \(\binom{8}{1}\).

   Thus, the total number of ways for this case is \(\binom{4}{4} \times \binom{8}{1} = 1 \times 8 = 8\).

3. 5 red marbles:

   The number of ways to choose 5 red marbles from 8 is \(\binom{8}{5}\).

   Thus, the total number of ways for this case is \(\binom{8}{5} = 56\).

Adding all these cases together, the total number of selections is:

\(280 + 8 + 56 = 344\).