Question

A dice marked with digit $\{1,2,2,3,3,3\}$, are thrown three times, then the probability of getting sum of the number on the face of the dice is six, is equal to-

• A. $\frac{7}{216}$
• B. $\frac{44}{216}$
• C. $\frac{14}{216}$
• D. $\frac{84}{216}$

Answer 1

The correct option is B $\frac{44}{216}$

$P\left(1\right)=\frac{1}{6}$, $P\left(2\right)=\frac{2}{6}$, $P\left(3\right)=\frac{3}{6}$

There are two cases possible:
$\{1,2,3\}\text{or}\{2,2,2\}$

Required probability $=(\frac{1}{6}\times \frac{2}{6}\times \frac{3}{6})\times 3!+(\frac{2}{6}\times \frac{2}{6}\times \frac{2}{6})=\frac{44}{216}$