Question

Three numbers a, b and c are selected independently and uniformly at random from the set {1, 2, 3}. Then the probability that these three numbers can form the sides of a triangle is

• A. $\frac{4}{27}$
• B. $\frac{2}{27}$
• C. $\frac{5}{9}$
• D. $\frac{1}{27}$

Answer 1

Answer: (C)

$\frac{5}{9}$

Answer 2

The total number of ways to choose three numbers $a, b, c$ independently and uniformly from $\{1, 2, 3\}$ is $3^3 = 27$. For $a, b, c$ to form a triangle, they must satisfy the triangle inequality: $a+b>c$, $a+c>b$, and $b+c>a$. The triplets that form a triangle are: (1,1,1) (1,2,2), (2,1,2), (2,2,1) (1,3,3), (3,1,3), (3,3,1) (2,2,2) (2,2,3), (2,3,2), (3,2,2) (2,3,3), (3,2,3), (3,3,2) (3,3,3) There are $1 + 3 + 3 + 1 + 3 + 3 + 1 = 15$ such triplets. The probability is $\frac{15}{27} = \frac{5}{9}$.