Question

A pair of dice is thrown thrice. The probability of throwing doublets at least once is

• A. $\frac{1}{36}$
• B. $\frac{25}{216}$
• C. $\frac{125}{216}$
• D. $\frac{91}{216}$

Answer 1

Answer: (D)

Doublets occur when the numbers thrown are $(1.1),(2,2),.....(6,6)$.
Therefore the probability of a doublet occurring in a single throw $=\frac{6}{36}=\frac{1}{6}$.
The probability of a doublet not occurring at all in three throws
$=\left(1-\frac{1}{6}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{6}\right)$
$={\left(\frac{5}{6}\right)}^3=\frac{125}{216}$.
$\therefore$ Required probability $=1-\frac{125}{216}=\frac{91}{216}$ .