You have 5 yellow (Y), 4 black (B), and 3 green (G) balls, so the total number of balls is: \[ 5 + 4 + 3 = 12 \text{ balls}. \] Non-black balls (yellow and green): \[ 5 \text{ yellow} + 3 \text{ green} = 8 \text{ non-black balls}. \] Now, if we are selecting 3 balls and the event is that no black ball is selected, we are only choosing from the 8 non-black balls. The probability of selecting 3 balls with no black balls is the ratio of the number of favorable outcomes (selecting 3 balls from the 8 non-black balls) to the total number of possible outcomes (selecting 3 balls from all 12 balls): \[ P(\text{No black ball is selected}) = \frac{\binom{8}{3}}{\binom{12}{3}}. \] Now, calculate the combinations: \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56, \] \[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220. \] Thus, the probability is: \[ P(\text{No black ball is selected}) = \frac{56}{220} = \frac{14}{55}. \]