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; then find the probability that the sum of the numbers appearing on the dice is $6$.

• A. $\frac{1}{3}$
• B. $\frac{5}{6}$
• C. $\frac{1}{6}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{6}$

The tetrahedron dice can have numbers from $1$ to $4$ only, while the common cubic dice can have from $1$ to $6$.

To get the sum of $6$, the tetrahedron dice should have numbers from $1$ to $4$ and the common cube should have $2$ to $5$ also.

There are four cases to get the sum of $6$ in (Tetrahedron, Common) as $(1,5),(2,4),(3,3)$ and $(4,2)$.

But total cases are $6\times 4$

Hence, the probability is $\frac{4}{6\times 4}=\frac{1}{6}$