Tamika selects two different numbers at random from the set $ \{ 8,9,10\} $ and adds them. Carlos takes two different numbers at random from the set $ \{ 3,5,6\} $ and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?

$ \text{(A)}\ \frac{4}{9}\qquad\text{(B)}\ \frac{5}{9}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{1}{3}\qquad\text{(E)}\ \frac{2}{3} $
1. **Identify the possible sums for Tamika:**
   Tamika selects two different numbers from the set \(\{8, 9, 10\}\) and adds them. The possible sums are:
   \[
   8 + 9 = 17, \quad 8 + 10 = 18, \quad 9 + 10 = 19
   \]
   Therefore, the possible sums are \(17, 18, 19\).

2. **Identify the possible products for Carlos:**
   Carlos selects two different numbers from the set \(\{3, 5, 6\}\) and multiplies them. The possible products are:
   \[
   3 \times 5 = 15, \quad 3 \times 6 = 18, \quad 5 \times 6 = 30
   \]
   Therefore, the possible products are \(15, 18, 30\).

3. **List all possible pairs \((\text{Tamika's sum}, \text{Carlos' product})\):**
   The pairs are:
   \[
   (17, 15), \quad (17, 18), \quad (17, 30), \quad (18, 15), \quad (18, 18), \quad (18, 30), \quad (19, 15), \quad (19, 18), \quad (19, 30)
   \]

4. **Count the pairs where Tamika's sum is greater than Carlos' product:**
   - \((17, 15)\): \(17 > 15\)
   - \((17, 18)\): \(17 \not> 18\)
   - \((17, 30)\): \(17 \not> 30\)
   - \((18, 15)\): \(18 > 15\)
   - \((18, 18)\): \(18 \not> 18\)
   - \((18, 30)\): \(18 \not> 30\)
   - \((19, 15)\): \(19 > 15\)
   - \((19, 18)\): \(19 > 18\)
   - \((19, 30)\): \(19 \not> 30\)

   The pairs where Tamika's sum is greater than Carlos' product are:
   \[
   (17, 15), \quad (18, 15), \quad (19, 15), \quad (19, 18)
   \]
   There are 4 such pairs.

5. **Calculate the probability:**
   There are a total of 9 possible pairs. The probability that Tamika's result is greater than Carlos' result is:
   \[
   \frac{\text{Number of favorable pairs}}{\text{Total number of pairs}} = \frac{4}{9}
   \]

The final answer is \(\boxed{\frac{4}{9}}\).