Problem 109

To prove: Since the possible values for the random variable $\xi$ representing the occurrence of event A are 0 and 1, we have:
$$P(\xi=0) = 1-p,$$
$$P(\xi=1) = p.$$

Thus, the expected value of $\xi$ is:
$$E(\xi) = 0 \cdot (1-p) + 1 \cdot p = p.$$

The variance of $\xi$, denoted as $D(\xi)$, is calculated as follows:
$$D(\xi) = (0-p)^2 \cdot (1-p) + (1-p)^2 \cdot p$$
$$= p(1-p) \leq \left(\frac{p+(1-p)}{2}\right)^2 = \frac{1}{4}.$$

The inequality used here is based on the fact that the expression $p(1-p)$ reaches its maximum when $p = \frac{1}{2}$, which can be understood from the basic inequality (since the arithmetic mean is always greater than or equal to the geometric mean):
$$\left(\frac{p+(1-p)}{2}\right)^2 \geq p(1-p),$$
which simplifies to:
$$D(\xi) \leq \frac{1}{4}.$$

Therefore, the maximum variance of the number of times event A occurs in one experiment, $X$, is $\boxed{\frac{1}{4}}$. Hence, the correct answer is C.

Enhancing the solution, we identify that $\xi$ is the random variable indicating the occurrence of event A in a single trial, taking the values 0 or 1. By expressing the variances formally, applying the basic inequality, and simplifying, we can determine the maximum possible value. This problem tests the application of variance for discrete random variables and the use of basic inequalities, integrating both probability and general mathematical knowledge into one comprehensive question.