Question

The probability of getting $10$ in a single throw of three fair dice is :

• A. $\frac{1}{6}$
• B. $\frac{1}{8}$
• C. $\frac{1}{9}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{8}$

Answer 2

Exhaustive no. of cases $=6^{3}$ $10$ can appear on three dice either as distinct number as following $(1,3,6) ;(1,4,5) ;(2,3,5)$ and each can occur in $3 !$ ways. Or $10$ can appear on three dice as repeated digits as following $(2,2,6),(2,4,4),(3,3,4)$ and each can occur in $\frac{3 !}{2 !}$ ways. $\therefore$ No. of favourable cases $=3 \times 3 !+3 \times \frac{3 !}{2 !}=27$