Three six-faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is `k(3 le k le 8)`, is

Options

A. `((k-1)(k-2))/(432)`
B. `(k(k-1))/(432)`
C. `(k^2)/(432)`
D. None of these

Answer

To solve the problem of finding the probability that the sum of the numbers appearing on three six-faced dice is \( k \) (where \( 3 \leq k \leq 8 \)), we can follow these steps:

### Step 1: Determine the Total Number of Outcomes
When three six-faced dice are thrown, each die has 6 faces. Therefore, the total number of outcomes when throwing three dice is:
\[ 6 \times 6 \times 6 = 216 \]

### Step 2: Define the Favorable Outcomes
We need to find the number of favorable outcomes where the sum of the numbers on the three dice equals \( k \) for \( k = 3, 4, 5, 6, 7, 8 \).

### Step 3: Use Generating Functions
The generating function for a single die is:
\[ x + x^2 + x^3 + x^4 + x^5 + x^6 = x \frac{1 - x^6}{1 - x} \]
For three dice, the generating function becomes:
\[ (x + x^2 + x^3 + x^4 + x^5 + x^6)^3 = \left(x \frac{1 - x^6}{1 - x}\right)^3 = x^3 \frac{(1 - x^6)^3}{(1 - x)^3} \]

### Step 4: Extract Coefficients
To find the coefficient of \( x^k \) in the expansion, we need to find the coefficient of \( x^{k-3} \) in:
\[ (1 - x^6)^3 (1 - x)^{-3} \]
Using the binomial theorem, we expand \( (1 - x^6)^3 \):
\[ (1 - x^6)^3 = \sum_{j=0}^{3} \binom{3}{j} (-1)^j x^{6j} = 1 - 3x^6 + 3x^{12} - x^{18} \]
The term \( (1 - x)^{-3} \) can be expanded using the generalized binomial series:
\[ (1 - x)^{-3} = \sum_{n=0}^{\infty} \binom{n + 2}{2} x^n \]

### Step 5: Combine the Expansions
To find the coefficient of \( x^{k-3} \), we need to consider the contributions from the expansion:
\[ \text{Coefficient of } x^{k-3} = \binom{k-3 + 2}{2} - 3 \cdot \binom{k-3 - 6 + 2}{2} \quad \text{(for } k \geq 9\text{)} \]

### Step 6: Calculate the Probability
The probability \( P \) that the sum of the numbers appearing on the dice is \( k \) is given by:
\[ P(k) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{\text{Coefficient of } x^{k-3}}{216} \]

### Step 7: Finalize the Results
Now we can compute the probabilities for \( k = 3, 4, 5, 6, 7, 8 \) using the coefficients derived from the above steps.