3 six faced dice are thrown. One die bears numbers 0 to 5, second bears numbers 1 to 6 and third has numbers 2 to 7. The probability that sum of numbers shown is six is

Options

A. `7/108`
B. `5/108`
C. `5/216`
D. `7/216`

Answer

To solve the problem of finding the probability that the sum of the numbers shown on three different dice is equal to 6, we can follow these steps:

### Step 1: Identify the Dice and Their Faces
- **Die 1**: Numbers from 0 to 5 (6 outcomes)
- **Die 2**: Numbers from 1 to 6 (6 outcomes)
- **Die 3**: Numbers from 2 to 7 (6 outcomes)

### Step 2: Calculate the Total Number of Outcomes
The total number of outcomes when throwing the three dice can be calculated by multiplying the number of outcomes for each die:
\[ \text{Total Outcomes} = 6 \times 6 \times 6 = 216 \]

### Step 3: Find Favorable Outcomes for the Sum of 6
We need to find all combinations of the numbers on the three dice that add up to 6. Let's denote the outcomes of the three dice as \(x_1\), \(x_2\), and \(x_3\) respectively, where:
- \(x_1\) can be \(0, 1, 2, 3, 4, 5\)
- \(x_2\) can be \(1, 2, 3, 4, 5, 6\)
- \(x_3\) can be \(2, 3, 4, 5, 6, 7\)

We need to find all combinations such that:
\[ x_1 + x_2 + x_3 = 6 \]

### Step 4: List the Combinations
We can systematically list the combinations:
1. \(x_1 = 0\): \(x_2 + x_3 = 6\) → (1,5), (2,4), (3,3), (4,2), (5,1) → 5 combinations
2. \(x_1 = 1\): \(x_2 + x_3 = 5\) → (1,4), (2,3), (3,2), (4,1) → 4 combinations
3. \(x_1 = 2\): \(x_2 + x_3 = 4\) → (1,3), (2,2), (3,1) → 3 combinations
4. \(x_1 = 3\): \(x_2 + x_3 = 3\) → (1,2), (2,1) → 2 combinations
5. \(x_1 = 4\): \(x_2 + x_3 = 2\) → (1,1) → 1 combination
6. \(x_1 = 5\): No valid combinations since \(x_2\) must be at least 1 and \(x_3\) must be at least 2.

### Step 5: Count the Total Favorable Outcomes
Adding the combinations:
- From \(x_1 = 0\): 5
- From \(x_1 = 1\): 4
- From \(x_1 = 2\): 3
- From \(x_1 = 3\): 2
- From \(x_1 = 4\): 1

Total favorable outcomes = \(5 + 4 + 3 + 2 + 1 = 15\)

### Step 6: Calculate the Probability
The probability \(P\) that the sum of the numbers shown is 6 can be calculated as:
\[ P = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes}} = \frac{15}{216} \]

### Step 7: Simplify the Probability
To simplify \(\frac{15}{216}\), we can divide both the numerator and the denominator by 3:
\[ P = \frac{5}{72} \]

### Final Answer
The probability that the sum of the numbers shown is 6 is \(\frac{5}{72}\).