Question

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

• A. $\frac{125}{216}$
• B. $\frac{126}{181}$
• C. $\frac{5}{36}$
• D. $\frac{25}{36}$

Answer 1

The correct option is D $\frac{25}{36}$

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