A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability $\mathrm{P}(\mathrm{X} \ge 5 \left| \mathrm{X}> 2\right| )$ is

from the given data $\mathrm{p}=\frac{1}{6}, \mathrm{q}=\frac{5}{6}$

 $$\begin{gathered} \mathrm{Now} \mathrm{P}(X\ge 5)=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^4\times \frac{5}{6}\times \frac{1}{6}+{\left(\frac{5}{6}\right)}^6 \left(\frac{1}{6}\right)+.........\infty \\ ={\left(\frac{5}{6}\right)}^4\times \frac{1}{6}\left[1+\frac{5}{6}+{\left(\frac{5}{6}\right)}^2+.......+\infty \right] \\ =\frac{5^4}{6^5}\times \left[\frac{1}{1-{\displaystyle \frac{5}{6}}}\right] \\ =\frac{5^4}{6^5}\times 6={\left(\frac{5}{6}\right)}^4 \end{gathered}$$and $\mathrm{P}(X>2)={\left(\frac{5}{6}\right)}^2\times \frac{1}{6}\times 6={\left(\frac{5}{6}\right)}^2$

Now $\mathrm{P}(X\ge 5/X>2)$

       $$\begin{gathered} =\frac{\mathrm{P}(X\ge 5\cap X>2)}{\mathrm{P}(X>2)} \\ =\frac{\mathrm{P}(X\ge 5)}{\mathrm{P}(X>2)} \\ =\frac{{\left(\frac{5}{6}\right)}^4}{{\left(\frac{5}{6}\right)}^2} \\ ={\left(\frac{5}{6}\right)}^2 \\ =\frac{25}{36} \end{gathered}$$