Question:

**Step 1: Count balls in each box** The contents of each box are: * Box $B_1$: 1 white ball, 3 red balls, 2 black balls. Total balls: $1+3+2=6$. * Box $B_2$: 2 white balls, 3 red balls, 4 black balls. Total balls: $2+3+4=9$. * Box $B_3$: 3 white balls, 4 red balls, 5 black balls. Total balls: $3+4+5=12$. **Step 2: Calculate the probability that all 3 drawn balls are white** The probability of drawing a white ball from each box is: $$P(\text{all White}) = P(W_1) \times P(W_2) \times P(W_3) = \frac{1}{6} \times \frac{2}{9} \times \frac{3}{12} = \frac{6}{648}$$ **Step 3: Calculate the probability that all 3 drawn balls are red** The probability of drawing a red ball from each box is: $$P(\text{all Red}) = P(R_1) \times P(R_2) \times P(R_3) = \frac{3}{6} \times \frac{3}{9} \times \frac{4}{12} = \frac{36}{648}$$ **Step 4: Calculate the probability that all 3 drawn balls are black** The probability of drawing a black ball from each box is: $$P(\text{all Black}) = P(B_1) \times P(B_2) \times P(B_3) = \frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} = \frac{40}{648}$$ **Step 5: Calculate the total probability that all 3 drawn balls are of the same color** The events of drawing all white, all red, or all black balls are mutually exclusive. Therefore, the total probability that all 3 drawn balls are of the same color is the sum of their individual probabilities: $$P(\text{same color}) = P(\text{all White}) + P(\text{all Red}) + P(\text{all Black})$$ $$P(\text{same color}) = \frac{6}{648} + \frac{36}{648} + \frac{40}{648}$$ $$P(\text{same color}) = \frac{6 + 36 + 40}{648} = \frac{82}{648}$$