Question

A person throws two dice, one the common cube, and the other a regular tetrahedron, the number on the lowest face being taken in the case of the tetrahedron; what is the chance that the sum of the numbers thrown is not less than $5$?

Answer 1

A cube dice has 6 sides and a tetrahedron dice has 4 sides

$\therefore$ the two dices may be thrown in $6\times 4=24$ ways

For the dices to get less than 5 in the following 6 ways : $(3,1),(2,2),(1,3),(2,1),(1,2),(1,1)$, where the first digit is for cube dice and the second for tetrahedron dice

$\therefore$ The probability of NOT getting a number less than 5 $=1-\frac{6}{24}=\frac{3}{4}$