Question

Find the chance of throwing $10$ exactly in one throw with $3$ dice.

Answer 1

With 3 dices 10 can be thrown in the follwing ways

1) $631$ - This can happen in $3!=6$ ways

2) $622$ - This can happen in $\frac{3!}{2!}=3$ ways

3) $541$ - This can happen in $3!=6$ ways

4) $532$ - This can happen in $3!=6$ ways

5) $442$ - This can happen in $\frac{3!}{2!}=3$ ways

6) $433$ - This can happen in $\frac{3!}{2!}=3$ ways

$\therefore$ Total number of favourable cases $=6+3+6+6+3+3=27$

Also, Total number of cases $=6^3=216$

$\therefore$ Required Probability $=\frac{27}{216}=\frac{1}{8}$