Probability That Die Tossed Until Six Requires Three Tosses

To find the probability that $X = 3$, we consider the geometric distribution model for the number of independent trials required to achieve the first success.

Let obtaining a six on a single roll of a fair die be defined as a "success" ($S$), and obtaining any other number (from $1$ to $5$) be defined as a "failure" ($F$).

The probability of success on any single toss is:
$P(S) = \frac{1}{6}$

The probability of failure on any single toss is:
$P(F) = 1 - P(S) = 1 - \frac{1}{6} = \frac{5}{6}$

The random variable $X$ denotes the number of tosses required until the first six appears. The event $X = 3$ means that:

1. The first toss results in a failure ($F$),
2. The second toss results in a failure ($F$),
3. The third toss results in a success ($S$).

Since the outcomes of successive tosses are independent, the probability is given by:
$P(X = 3) = P(F) \times P(F) \times P(S)$

Substituting the respective probabilities:
$P(X = 3) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) = \frac{25}{216}$

Thus, the correct option is A.