Step-by-step Solution:

1. Define the Random Variable and its Distribution Let $X$ be the random variable representing the number of tosses required to obtain the first six. The experiment consists of a sequence of independent Bernoulli trials, where success is getting a '6' and failure is not getting a '6'. This means $X$ follows a geometric distribution.

2. Identify Probabilities of Success and Failure The die is fair, so there are 6 equally likely outcomes for each toss.

   • The probability of success (getting a six) in a single toss is $p = P(6) = \frac{1}{6}$.
   • The probability of failure (not getting a six) in a single toss is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.

3. State the Probability Mass Function (PMF) The probability of getting the first six on the $k$-th toss is given by the PMF of the geometric distribution: $P(X=k) = q^{k-1}p = \left(\frac{5}{6}\right)^{k-1} \left(\frac{1}{6}\right)$

4. Formulate the Conditional Probability We need to find the conditional probability that $X \ge 6$ given that $X > 3$, which is denoted by $P(X \ge 6 | X > 3)$. The formula for conditional probability is: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ In our case, event $A$ is $X \ge 6$ and event $B$ is $X > 3$.

5. Determine the Intersection of the Events The event $A$ is $X \in \{6, 7, 8, ...\}$. The event $B$ is $X \in \{4, 5, 6, ...\}$. The intersection $A \cap B$ is the set of outcomes common to both events, which is $X \in \{6, 7, 8, ...\}$. Therefore, the intersection event $A \cap B$ is the same as event $A$, i.e., $X \ge 6$.

6. Rewrite the Conditional Probability Formula Substituting the intersection back into the formula, we get: $P(X \ge 6 | X > 3) = \frac{P(X \ge 6)}{P(X > 3)}$

7. Calculate the Required Probabilities We need to calculate $P(X > 3)$ and $P(X \ge 6)$.

   • The event $X > 3$ means that the first three tosses did not result in a six. The probability of this is: $P(X > 3) = q \times q \times q = q^3 = \left(\frac{5}{6}\right)^3$

   • The event $X \ge 6$ means that the first five tosses did not result in a six. (If the first six appears on toss 6 or later, the first 5 must be failures). The probability of this is: $P(X \ge 6) = q \times q \times q \times q \times q = q^5 = \left(\frac{5}{6}\right)^5$

8. Compute the Final Conditional Probability Now, we substitute the probabilities from Step 7 into the formula from Step 6: $P(X \ge 6 | X > 3) = \frac{\left(\frac{5}{6}\right)^5}{\left(\frac{5}{6}\right)^3}$ $P(X \ge 6 | X > 3) = \left(\frac{5}{6}\right)^{5-3} = \left(\frac{5}{6}\right)^2$ $P(X \ge 6 | X > 3) = \frac{25}{36}$

Alternative Method (using Memoryless Property):

The geometric distribution has a memoryless property, which states that $P(X > s+t | X > s) = P(X > t)$.

Let's apply this to our problem. We want to find $P(X \ge 6 | X > 3)$. This is equivalent to $P(X > 5 | X > 3)$. Let $s=3$ and $t=2$. Then $s+t=5$. So, $P(X > 3+2 | X > 3) = P(X > 2)$.

$P(X > 2)$ is the probability that the first two tosses are not a six. $P(X > 2) = q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

Both methods give the same result.

Conclusion:

The conditional probability that $X \ge 6$ given $X > 3$ is $\frac{25}{36}$. Comparing this with the given options:

A: $\frac{125}{216}$

B: $\frac{25}{216}$

C: $\frac{5}{36}$

D: $\frac{25}{36}$

The correct option is D.