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?

• A. $\frac{18}{24}=\frac{3}{4}$
• B. $\frac{6}{24}=\frac{1}{4}$
• C. $\frac{6}{18}=\frac{1}{3}$
• D. $\frac{12}{18}=\frac{2}{3}$

Answer 1

Answer: (A)

$\frac{18}{24}=\frac{3}{4}$

Since, a dice has six faces and a regular tetrahedron has $4$ faces.
So, total number of cases $=6\times 4=24$
Now, for sum S, $5\le S\le 10$
Favorable cases $\left\{\right(1,4\left)\right(2,4\left)\right(3,4),(4,4),(5,4),(6,4),(6,3),(5,3),(4,3),(3,3),(2,3\left)\right(6,2),(5,2),(4,2),(3,2),(6,1),(5,1),(4,1\left)\right\}$
Number of favorable cases $=18$
Hence, probability $=\frac{18}{24}=\frac{3}{4}$