Use multinomial distribution with $p_i = 1/8$, $n=93, n_1 = 33, n_2 = 7, n_3 = 8, n_4 = 1, n_5 = 2, n_6 = 0, n_7 = 5, n_8 = 37$ for i = 1 to 8. The required prob = $$\frac{93!}{33!\cdot7!\cdot8!\cdot1!\cdot2!\cdot0!\cdot5!\cdot37!}\cdot{(\frac{1}{8})}^{93}$$ <