The rule for winning the game is changed. The game is now won if the product of the numbers, recorded in the table, from one throw of the two dice is \( n \), where \( n \in \mathbb{N} \). The probability of winning the game is now \( \frac{1}{36} \). List the possible values of \( n \).

To determine the possible values of \( n \) that result in a winning probability of \( \frac{1}{36} \), we consider the following: 1. **Assumption**: We're using two standard six-sided dice, each showing numbers from 1 to 6. 2. **Total Outcomes**: There are \( 6 \times 6 = 36 \) possible outcomes when rolling two dice. 3. **Winning Condition**: The game is won if the product of the numbers on the two dice equals \( n \). For the probability to be \( \frac{1}{36} \), there must be exactly **one** pair \((a, b)\) such that \( a \times b = n \). 4. **Possible Products and Their Frequencies**: - **Unique Products**: These are products that can only be achieved in one way. - \( 1 = 1 \times 1 \) - \( 9 = 3 \times 3 \) - \( 16 = 4 \times 4 \) - \( 25 = 5 \times 5 \) - \( 36 = 6 \times 6 \) - **Non-Unique Products**: These can be achieved in multiple ways (e.g., \( 2 = 1 \times 2 \) and \( 2 \times 1 \)), hence their probabilities would be greater than \( \frac{1}{36} \). 5. **Conclusion**: The values of \( n \) that satisfy the winning condition with a probability of \( \frac{1}{36} \) are those with exactly one corresponding outcome. **Possible values of \( n \)**: \[ 1, \ 9, \ 16, \ 25, \ 36 \] **Answer:** All positive integers equal to 1, 9, 16, 25, or 36