Step-by-step Solution:

We are given the parameters of a binomial distribution and need to find the probability of success, \(p\).

1. List the given information and relevant formulas:
Number of trials, \(n = 9\).
Variance, \(\text{Var}(X) = 2\).
Probability of success, \(p\).
Probability of failure, \(q = 1 - p\).
The formula for the variance of a binomial distribution is \(\text{Var}(X) = npq\).
A key condition is that the probability of success is greater than the probability of failure, i.e., \(p > q\).

2. Set up and solve the equation for p:
Substitute the given values into the variance formula:
\[2 = 9 \cdot p \cdot q\]
Now, replace \(q\) with \(1-p\) to create an equation with only one variable, \(p\):
\[2 = 9p(1 - p)\]
\[2 = 9p - 9p^2\]
Rearrange this into the standard quadratic form (\(ax^2+bx+c=0\)):
\[9p^2 - 9p + 2 = 0\]
3. Solve the quadratic equation:
This equation can be solved by factoring. We need two numbers that multiply to \(9 \times 2 = 18\) and add to \(-9\). These numbers are \(-6\) and \(-3\).
\[\begin{aligned} 9p^2 - 6p - 3p + 2 &= 0 \\ 3p(3p - 2) - 1(3p - 2) &= 0 \\ (3p - 1)(3p - 2) &= 0 \end{aligned}\]
This gives two possible solutions for \(p\):
Solution 1: \(3p - 1 = 0 \implies p = \frac{1}{3}\)
Solution 2: \(3p - 2 = 0 \implies p = \frac{2}{3}\)

4. Apply the condition \(p > q\) to find the correct solution:
We must check which of these solutions satisfies the given condition that success is more probable than failure.
Case 1: If \(p = \frac{1}{3}\), then \(q = 1 - \frac{1}{3} = \frac{2}{3}\). In this case, \(p < q\), which violates the condition.
Case 2: If \(p = \frac{2}{3}\), then \(q = 1 - \frac{2}{3} = \frac{1}{3}\). In this case, \(p > q\), which satisfies the condition.

Result:
The correct probability of success is \(\frac{2}{3}\).