A fair die is tossed until six is obtained on it. Let \( X \) be the number of required tosses, then the conditional probability \( P(X \ge 5 \mid X > 2) \) is:
\( \dfrac{11}{36} \)
\( \dfrac{25}{36} \)
\( \dfrac{5}{6} \)
\( \dfrac{125}{216} \)