aime26-08

Proposed solution

To solve this problem, we need to determine the conditional probability that exactly one face of the die remains blank given that all even-numbered stickers (2, 4, and 6) are visible. Let's start by analyzing the situation step by step.

First, we note that Joanne has six stickers labeled 1 through 6. She rolls the die and places each sticker on top, covering any existing sticker. We need to find the probability that exactly one face remains blank given that all even-numbered stickers (2, 4, and 6) are visible.

### Step 1: Total Possible Outcomes
Joanne rolls the die six times, each time placing a sticker on top. The total number of ways to place the stickers is \(6!\) (factorial of 6), which is:
\[
6! = 720
\]

### Step 2: Favorable Outcomes
We need to count the number of ways to place the stickers such that exactly one face remains blank and all even-numbered stickers (2, 4, and 6) are visible.

1. **Choosing the blank face**: There are 6 faces on the die, so there are 6 ways to choose which face will remain blank.
2. **Placing the stickers**: Once we choose the blank face, we need to place the stickers 1, 3, 5, 2, 4, and 6 on the remaining 5 faces. The stickers 2, 4, and 6 must be visible, and we need to ensure they are not on the blank face.

Let's denote the blank face as \(B\). The remaining faces are \(A, C, D, E, F\). We need to place 1, 3, 5, 2, 4, and 6 on these 5 faces.

### Step 3: Arranging the Stickers
We need to place 1, 3, 5, 2, 4, and 6 on the 5 faces such that 2, 4, and 6 are visible. This means that 2, 4, and 6 must not be on \(B\). We can place 1, 3, and 5 on the remaining 4 faces, and then place 2, 4, and 6 on the remaining 3 faces.

1. **Choosing 3 faces for 1, 3, and 5**: There are \(\binom{4}{3}\) ways to choose 3 faces out of 4 for 1, 3, and 5.
\[
\binom{4}{3} = 4
\]
2. **Arranging 1, 3, and 5 on these 3 faces**: There are \(3!\) ways to arrange 1, 3, and 5 on these 3 faces.
\[
3! = 6
\]
3. **Arranging 2, 4, and 6 on the remaining 3 faces**: There are \(3!\) ways to arrange 2, 4, and 6 on these 3 faces.
\[
3! = 6
\]

Thus, for each choice of the blank face, there are:
\[
4 \times 6 \times 6 = 144
\]
ways to place the stickers such that exactly one face remains blank and all even-numbered stickers are visible.

### Step 4: Total Favorable Outcomes
Since there are 6 choices for the blank face, the total number of favorable outcomes is:
\[
6 \times 144 = 864
\]

### Step 5: Conditional Probability
We need to find the conditional probability given that all even-numbered stickers are visible. We need to count the total number of outcomes where all even-numbered stickers (2, 4, and 6) are visible.

1. **Choosing 3 faces for 2, 4, and 6**: There are \(\binom{6}{3}\) ways to choose 3 faces out of 6 for 2, 4, and 6.
\[
\binom{6}{3} = 20
\]
2. **Arranging 2, 4, and 6 on these 3 faces**: There are \(3!\) ways to arrange 2, 4, and 6 on these 3 faces.
\[
3! = 6
\]
3. **Arranging 1, 3, and 5 on the remaining 3 faces**: There are \(3!\) ways to arrange 1, 3, and 5 on these 3 faces.
\[
3! = 6
\]

Thus, the total number of outcomes where all even-numbered stickers are visible is:
\[
20 \times 6 \times 6 = 720
\]

### Step 6: Calculating the Conditional Probability
The conditional probability \(p\) that exactly one face remains blank given that all even-numbered stickers are visible is:
\[
p = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes where all even-numbered stickers are visible}} = \frac{864}{720} = \frac{6}{5}
\]

### Step 7: Simplifying the Fraction
The fraction \(\frac{6}{5}\) is already in its simplest form. Therefore, \(m = 6\) and \(n = 5\).

### Final Answer
\[
m + n = 6 + 5 = 11
\]

Thus, the final answer is:
\[
\boxed{11}
\]