For 600 throws, I would use a normal approximation with μ = (1/6)×600 = 100 and σ² = (1/6)×(1–1/6)×600 = 250/3 to calculate P(89.5 < X < 100.5) and P(119.5 < X < 120.5), especially since these values are quite close to the expected mean μ = 100.

Of course, you could use an infinite-precision package like Matlab or Mathematica to calculate the exact answer.