To find the probability that the sum of the numbers appearing on three dice is 9, we can follow these steps:
### Step 1: Determine the Total Number of Outcomes
When a single die is thrown, there are 6 possible outcomes (1 through 6). When three dice are thrown together, the total number of outcomes is calculated as:
\[
\text{Total outcomes} = 6 \times 6 \times 6 = 6^3 = 216
\]
### Step 2: Find the Favorable Outcomes for a Sum of 9
We need to find the number of combinations of the three dice that result in a sum of 9. We can do this systematically by considering the possible values for each die.
Let the three dice be represented as \(d_1\), \(d_2\), and \(d_3\). We need to find the number of non-negative integer solutions to the equation:
\[
d_1 + d_2 + d_3 = 9
\]
with the constraints \(1 \leq d_1, d_2, d_3 \leq 6\).
### Step 3: Change of Variables
To simplify the problem, we can change the variables:
Let \(d_1' = d_1 - 1\), \(d_2' = d_2 - 1\), \(d_3' = d_3 - 1\). This transforms our equation to:
\[
d_1' + d_2' + d_3' = 6
\]
where \(0 \leq d_1', d_2', d_3' \leq 5\).
### Step 4: Use the Stars and Bars Method
The number of non-negative integer solutions to the equation \(d_1' + d_2' + d_3' = 6\) without any upper limit is given by the stars and bars theorem:
\[
\text{Number of solutions} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28
\]
### Step 5: Subtract Invalid Cases
Now, we need to exclude cases where any of \(d_1', d_2', d_3'\) exceeds 5. If, for example, \(d_1' > 5\), we can set \(d_1'' = d_1' - 6\) (which is non-negative) and rewrite the equation:
\[
d_1'' + d_2' + d_3' = 0
\]
This has only one solution: \(d_1'' = 0\), \(d_2' = 0\), \(d_3' = 0\) (i.e., \(d_1 = 6\), \(d_2 = 1\), \(d_3 = 1\) or permutations thereof). There are 3 such cases (one for each die).
### Step 6: Calculate the Valid Outcomes
Thus, the number of favorable outcomes for the sum of 9 is:
\[
\text{Favorable outcomes} = 28 - 3 = 25
\]
### Step 7: Calculate the Probability
Finally, the probability \(P\) that the sum of the numbers appearing on the three dice is 9 is given by:
\[
P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{25}{216}
\]
### Conclusion
Thus, the probability that the sum of the numbers appearing on three dice is 9 is:
\[
\frac{25}{216}
\]
