Question

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required.
The conditional probability that $X \geq 6$ given $X>3$ equals

• A. $\frac{125}{216}$
• B. $\frac{25}{216}$
• C. $\frac{5}{36}$
• D. $\frac{25}{36}$

Answer 1

Answer: (D)

For $X \geq 6$, the probability is
$\frac{5^{5}}{6^{6}}+\frac{5^{6}}{6^{7}}+\ldots \infty=\frac{5^{5}}{6^{6}}\left(\frac{1}{1-5 / 6}\right)=\left(\frac{5}{6}\right)^{5}$
For $X >3$
$\frac{5^{3}}{6^{4}}+\frac{5^{4}}{6^{5}}+\frac{5^{5}}{6^{6}}+\ldots \infty=\left(\frac{5}{6}\right)^{3}$
Hence the conditional probability $\frac{(5 / 6)^{6}}{(5 / 6)^{3}}=\frac{25}{36}$.