A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls and 4 black balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls.
If 1 ball is drawn from each of the boxes `B_(1),B_(2) and B_(3),` the probability that all 3 drawn balls are of the same color is

Options

A. `82//648`
B. `90//648`
C. `558//648`
D. `566//648`

Explanation

To solve the problem, we need to calculate the probability that all three drawn balls from boxes \( B_1 \), \( B_2 \), and \( B_3 \) are of the same color. We will break this down into steps.

### Step 1: Identify the total number of balls in each box
- **Box \( B_1 \)**: Contains 1 white, 3 red, and 2 black balls.
- Total = \( 1 + 3 + 2 = 6 \) balls.

- **Box \( B_2 \)**: Contains 2 white, 3 red, and 4 black balls.
- Total = \( 2 + 3 + 4 = 9 \) balls.

- **Box \( B_3 \)**: Contains 3 white, 4 red, and 5 black balls.
- Total = \( 3 + 4 + 5 = 12 \) balls.

### Step 2: Calculate the probabilities for each color
1. **Probability that all drawn balls are white**:
- From \( B_1 \): Probability of drawing a white ball = \( \frac{1}{6} \)
- From \( B_2 \): Probability of drawing a white ball = \( \frac{2}{9} \)
- From \( B_3 \): Probability of drawing a white ball = \( \frac{3}{12} = \frac{1}{4} \)

Therefore, the combined probability for all white balls:
\[ P(\text{all white}) = \frac{1}{6} \times \frac{2}{9} \times \frac{1}{4} = \frac{1 \times 2 \times 1}{6 \times 9 \times 4} = \frac{2}{216} = \frac{1}{108} \]

2. **Probability that all drawn balls are red**:
- From \( B_1 \): Probability of drawing a red ball = \( \frac{3}{6} = \frac{1}{2} \)
- From \( B_2 \): Probability of drawing a red ball = \( \frac{3}{9} = \frac{1}{3} \)
- From \( B_3 \): Probability of drawing a red ball = \( \frac{4}{12} = \frac{1}{3} \)

Therefore, the combined probability for all red balls:
\[ P(\text{all red}) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{3} = \frac{1 \times 1 \times 1}{2 \times 3 \times 3} = \frac{1}{18} \]

3. **Probability that all drawn balls are black**:
- From \( B_1 \): Probability of drawing a black ball = \( \frac{2}{6} = \frac{1}{3} \)
- From \( B_2 \): Probability of drawing a black ball = \( \frac{4}{9} \)
- From \( B_3 \): Probability of drawing a black ball = \( \frac{5}{12} \)

Therefore, the combined probability for all black balls:
\[ P(\text{all black}) = \frac{1}{3} \times \frac{4}{9} \times \frac{5}{12} = \frac{1 \times 4 \times 5}{3 \times 9 \times 12} = \frac{20}{324} = \frac{5}{81} \]

### Step 3: Add the probabilities for all same color
Now, we add the probabilities of all three cases:
\[ P(\text{all same color}) = P(\text{all white}) + P(\text{all red}) + P(\text{all black}) \]
\[ = \frac{1}{108} + \frac{1}{18} + \frac{5}{81} \]

To add these fractions, we need a common denominator. The least common multiple of \( 108, 18, \) and \( 81 \) is \( 108 \).

- Convert \( \frac{1}{18} \) to \( \frac{6}{108} \)
- Convert \( \frac{5}{81} \) to \( \frac{40}{108} \)

Now, we can add:
\[ P(\text{all same color}) = \frac{1}{108} + \frac{6}{108} + \frac{40}{108} = \frac{47}{108} \]

### Final Answer
The probability that all three drawn balls are of the same color is:
\[ \frac{47}{108} \]