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

Answer 1

Answer: (B)

$\frac{25}{36}$

$P(X\le 3)=1-\left[P\right(X=1)+P(X=2\left)\right]$

$=1-\left[\frac{1}{6}+\frac{1}{6}\left(\frac{5}{6}\right)\right]$

$=1-\frac{11}{36}=\frac{25}{36}$