Question

A fair die is rolled repeatedly until a six is obtained. Let X denote the number of rolls 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)

$\frac{25}{36}$

Let $X$ denote the number of rolls required.

For
$P(X\ge 6)$

$=P\left(6\right)+P\left(7\right)+P\left(8\right)...\infty$

$=\frac{5^5}{6^6}+\frac{5^6}{6^7}+\frac{5^7}{6^8}+....\infty$
This is an infinite G.P with common ratio as $\frac{5}{6}$
Hence
$Sum=S$

$=\frac{5^5}{6^6}\left(\frac{6}{6-5}\right)$
$=\frac{5^5}{6^5}$ ...$\left(i\right)$

For
$P(X>3)$

$P\left(4\right)+P\left(5\right)+...\infty$

$=\frac{5^3}{6^4}+\frac{5^4}{6^5}+\frac{5^5}{6^6}+....\infty$
$=\frac{5^3}{6^3}$ ...$\left(ii\right)$

Hence applying conditional probability
Equation (i) divide by (ii), we get

$\text{Required probability}=\frac{P(X\ge 6)}{P(X>3)}=\frac{\frac{5^5}{6^5}}{\frac{5^3}{6^3}}$
$=\frac{25}{36}$