Question

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

• A. $\frac{5}{6}$
• B. $\frac{125}{216}$
• C. $\frac{11}{36}$
• D. $\frac{25}{36}$

Answer 1

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

$P(X\ge 5|X>2)=\frac{P(X\ge 5\cap X>2)}{P(X>2)}$
$=\frac{P(X\ge 5)}{P(X>2)}$
Now,
$P(X\ge 5)=\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^4\frac{5}{6}\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^6\frac{1}{6}+\cdots \infty$
and $P(X>2)=\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+{\left(\frac{5}{6}\right)}^3\frac{1}{6}+{\left(\frac{5}{6}\right)}^4\frac{1}{6}+\cdots \infty$
$\therefore P(X\ge 5)={\left(\frac{5}{6}\right)}^4\cdot \frac{1}{6}(1+\frac{5}{6}+{\left(\frac{5}{6}\right)}^2+\cdots \infty )$
$=\frac{5^4}{6^5}\cdot 6={\left(\frac{5}{6}\right)}^4$
and $P(X>2)={\left(\frac{5}{6}\right)}^2\cdot \frac{1}{6}\cdot 6={\left(\frac{5}{6}\right)}^2$

$\therefore$ Required probability $=\frac{(5/6{)}^4}{(5/6{)}^2}=\frac{25}{36}$