A box `B_1` contains 1 white ball, 3 red balls and 2 black balls. Another 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`, then the probability that all 3 drawn balls are of the same colour , is

Options

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

Answer

To find the probability that all three drawn balls from boxes \( B_1 \), \( B_2 \), and \( B_3 \) are of the same color, we can follow these steps:

### Step 1: Determine the total number of balls in each box
- **Box \( B_1 \)**:
- White balls = 1
- Red balls = 3
- Black balls = 2
- Total = \( 1 + 3 + 2 = 6 \)

- **Box \( B_2 \)**:
- White balls = 2
- Red balls = 3
- Black balls = 4
- Total = \( 2 + 3 + 4 = 9 \)

- **Box \( B_3 \)**:
- White balls = 3
- Red balls = 4
- Black balls = 5
- Total = \( 3 + 4 + 5 = 12 \)

### Step 2: Calculate the probability of drawing three balls of the same color
The probability that all three balls drawn are of the same color can be calculated as the sum of the probabilities of drawing three white balls, three red balls, and three black balls.

#### Probability of drawing three white balls:
- From \( B_1 \): \( P(W_1) = \frac{1}{6} \)
- From \( B_2 \): \( P(W_2) = \frac{2}{9} \)
- From \( B_3 \): \( P(W_3) = \frac{3}{12} = \frac{1}{4} \)

So,
\[ P(\text{all white}) = P(W_1) \times P(W_2) \times P(W_3) = \frac{1}{6} \times \frac{2}{9} \times \frac{1}{4} = \frac{1 \times 2 \times 1}{6 \times 9 \times 4} = \frac{2}{216} = \frac{1}{108} \]

#### Probability of drawing three red balls:
- From \( B_1 \): \( P(R_1) = \frac{3}{6} = \frac{1}{2} \)
- From \( B_2 \): \( P(R_2) = \frac{3}{9} = \frac{1}{3} \)
- From \( B_3 \): \( P(R_3) = \frac{4}{12} = \frac{1}{3} \)

So,
\[ P(\text{all red}) = P(R_1) \times P(R_2) \times P(R_3) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{3} = \frac{1 \times 1 \times 1}{2 \times 3 \times 3} = \frac{1}{18} \]

#### Probability of drawing three black balls:
- From \( B_1 \): \( P(B_1) = \frac{2}{6} = \frac{1}{3} \)
- From \( B_2 \): \( P(B_2) = \frac{4}{9} \)
- From \( B_3 \): \( P(B_3) = \frac{5}{12} \)

So,
\[ P(\text{all black}) = P(B_1) \times P(B_2) \times P(B_3) = \frac{1}{3} \times \frac{4}{9} \times \frac{5}{12} = \frac{1 \times 4 \times 5}{3 \times 9 \times 12} = \frac{20}{324} = \frac{5}{81} \]

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

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

Converting each fraction:
- \( \frac{1}{108} = \frac{1}{108} \)
- \( \frac{1}{18} = \frac{6}{108} \)
- \( \frac{5}{81} = \frac{5 \times 4}{81 \times 4} = \frac{20}{108} \)

Now, summing them:
\[ P(\text{same color}) = \frac{1}{108} + \frac{6}{108} + \frac{20}{108} = \frac{1 + 6 + 20}{108} = \frac{27}{108} = \frac{1}{4} \]

### Final Answer
The probability that all three drawn balls are of the same color is \( \frac{1}{4} \).

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