Probability of Drawing Balls of Same Color From Three Boxes

To determine the probability that all three selected balls are of the same color, we consider the contents of each box and total number of balls in each:

1. Box $B_1$:

   • White balls: $1$
   • Red balls: $3$
   • Black balls: $2$
   • Total balls in $B_1 = 1 + 3 + 2 = 6$

2. Box $B_2$:

   • White balls: $2$
   • Red balls: $3$
   • Black balls: $4$
   • Total balls in $B_2 = 2 + 3 + 4 = 9$

3. Box $B_3$:

   • White balls: $3$
   • Red balls: $4$
   • Black balls: $5$
   • Total balls in $B_3 = 3 + 4 + 5 = 12$

The total number of possible outcomes when selecting one ball from each box is given by: $\text{Total Outcomes} = 6 \times 9 \times 12 = 648$

For the three balls to be of the same color, they can either all be White, all be Red, or all be Black. Since these three cases are mutually exclusive, we can sum their individual probabilities.

Case 1: All 3 balls are White

The probability of drawing a white ball from $B_1$, $B_2$, and $B_3$ is: $P(\text{3 White}) = \left(\frac{1}{6}\right) \times \left(\frac{2}{9}\right) \times \left(\frac{3}{12}\right) = \frac{1 \times 2 \times 3}{648} = \frac{6}{648}$

Case 2: All 3 balls are Red

The probability of drawing a red ball from $B_1$, $B_2$, and $B_3$ is: $P(\text{3 Red}) = \left(\frac{3}{6}\right) \times \left(\frac{3}{9}\right) \times \left(\frac{4}{12}\right) = \frac{3 \times 3 \times 4}{648} = \frac{36}{648}$

Case 3: All 3 balls are Black

The probability of drawing a black ball from $B_1$, $B_2$, and $B_3$ is: $P(\text{3 Black}) = \left(\frac{2}{6}\right) \times \left(\frac{4}{9}\right) \times \left(\frac{5}{12}\right) = \frac{2 \times 4 \times 5}{648} = \frac{40}{648}$

Total Probability

Adding the probabilities of all three mutually exclusive cases: $P(\text{Same Colour}) = P(\text{3 White}) + P(\text{3 Red}) + P(\text{3 Black})$ $P(\text{Same Colour}) = \frac{6}{648} + \frac{36}{648} + \frac{40}{648} = \frac{6 + 36 + 40}{648} = \frac{82}{648}$

Thus, the correct option is A.