Question

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

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

Answer 1

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

Given that
$P(X\le 2)=\frac{1}{6}+\frac{5}{6}\times \frac{1}{6}=\frac{11}{36}$
Hence, the required probability is $1-\left(\frac{11}{36}\right)=\frac{25}{36}$