Question

Find the chance of throwing more than $15$ in one throw with $3$ dice.

Answer 1

A throw amounting to $18$ must be made up of $6,6,6,$ and this can occur in $1$ way$;$ $17$ can be made up of $6,6,5$ which can occur in $3$ ways$;$ $16$ may be made up of $6,6,4$ and $6,5,5,$ each of which arranagements can occur in $3$ ways.
Therefore the number of favourable cases is $1+3+3+3,$ or $10$.
And the total number of cases is $6^3,$ or $216;$
Therefore the required chance $=\frac{10}{216}=\frac{5}{108}$.