A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the

$\mathrm{P}(\mathrm{x} \geq 5 \mid \mathrm{x}>2)=\frac{\mathrm{P}(\mathrm{x} \geq 5)}{\mathrm{P}(\mathrm{x}>2)}$

$\frac{\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}+\ldots \ldots+\infty}{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\ldots \ldots+\infty}$

$\frac{\frac{\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}}{1-\frac{5}{6}}}{\frac{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}}{1-\frac{5}{6}}}=\left(\frac{5}{6}\right)^{2}=\frac{25}{36}$