Question

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required.
The conditional probability that $X\ge 6$ 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 $X\ge 6$, the probability is
$\frac{5^5}{6^6}+\frac{5^6}{6^7}+\dots \infty =\frac{5^5}{6^6}\left(\frac{1}{1-5/6}\right)={\left(\frac{5}{6}\right)}^5$
For $X>3$
$\frac{5^3}{6^4}+\frac{5^4}{6^5}+\frac{5^5}{6^6}+\dots \infty ={\left(\frac{5}{6}\right)}^3$
Hence the conditional probability $\frac{(5/6{)}^6}{(5/6{)}^3}=\frac{25}{36}$.