Question:
A fair die is tossed until six is obtained on it. Let $\mathrm{X}$ be the number of required tosses, then the
conditional probability $\mathrm{P}(\mathrm{X} \geq 5 \mid \mathrm{X}>2)$ is :
Correct Option: , 4
Solution:
$\mathrm{P}(\mathrm{x} \geq 5 \mid \mathrm{x}>2)=\frac{\mathrm{P}(\mathrm{x} \geq 5)}{\mathrm{P}(\mathrm{x}>2)}$
$\frac{\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}+\ldots \ldots+\infty}{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\ldots \ldots+\infty}$
$\frac{\frac{\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}}{1-\frac{5}{6}}}{\frac{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}}{1-\frac{5}{6}}}=\left(\frac{5}{6}\right)^{2}=\frac{25}{36}$