To solve the problem, we need to find the value of \( x \) given the probabilities associated with the outcomes of a dice roll.
1. **Understanding the probabilities**:
- The probability of one number on the dice is \( \frac{1}{6} + x \).
- The probability of the number opposite to it is \( \frac{1}{6} - x \).
- The probabilities of the remaining four numbers are \( \frac{1}{6} \) each.
2. **Setting up the equation**:
- The total probability must equal 1:
\[
\left(\frac{1}{6} + x\right) + \left(\frac{1}{6} - x\right) + 4 \cdot \frac{1}{6} = 1
\]
- Simplifying this:
\[
\frac{1}{6} + x + \frac{1}{6} - x + \frac{4}{6} = 1
\]
\[
1 = 1
\]
- This confirms that our probabilities are valid.
3. **Calculating the probability of rolling a sum of 7**:
- The possible pairs that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
- The probabilities for these pairs are:
- For (1, 6): \( \left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) \)
- For (2, 5): \( \left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right) \)
- For (3, 4): \( \left(\frac{1}{6}\right)\left(\frac{1}{6}\right) \)
- For (4, 3): \( \left(\frac{1}{6}\right)\left(\frac{1}{6}\right) \)
- For (5, 2): \( \left(\frac{1}{6} - x\right)\left(\frac{1}{6} + x\right) \)
- For (6, 1): \( \left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) \)
4. **Summing the probabilities**:
- The total probability of rolling a sum of 7 is:
\[
2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) + 2\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)
\]
- This can be simplified:
\[
2\left(\frac{1}{36} - x^2\right) + 2\left(\frac{1}{36}\right) = \frac{2}{36} - 2x^2 + \frac{2}{36} = \frac{4}{36} - 2x^2 = \frac{1}{9} - 2x^2
\]
5. **Setting the equation equal to the given probability**:
- We know this probability equals \( \frac{13}{96} \):
\[
\frac{1}{9} - 2x^2 = \frac{13}{96}
\]
6. **Finding a common denominator**:
- The common denominator between 9 and 96 is 288:
\[
\frac{32}{288} - 2x^2 = \frac{39}{288}
\]
7. **Solving for \( x^2 \)**:
- Rearranging gives:
\[
-2x^2 = \frac{39}{288} - \frac{32}{288}
\]
\[
-2x^2 = \frac{7}{288}
\]
\[
2x^2 = -\frac{7}{288} \implies x^2 = \frac{7}{576}
\]
\[
x = \pm \sqrt{\frac{7}{576}} = \pm \frac{\sqrt{7}}{24}
\]
8. **Choosing the positive value**:
- Since probabilities cannot be negative, we take:
\[
x = \frac{\sqrt{7}}{24}
\]
Thus, the value of \( x \) is \( \frac{\sqrt{7}}{24} \).
