Question

Out of 3n consecutive integers, three are selected at random. The probability that their sum is divisible by 3 is

• A. $\frac{(n-1(n-2\left)\right)}{(3n-1)(3n-2)}$
• B. $\frac{2n^2-3n+2}{(3n-1)(3n-2)}$
• C. $\frac{2n^2}{(3n-1)(3n-2)}$
• D. $\frac{3n^2-3n+2}{(3n-1)(3n-2)}$

Answer 1

The correct option is D $\frac{3n^2-3n+2}{(3n-1)(3n-2)}$

Let $r,r+1,r+2,\dots \dots r+3n-1$ be 3n consecutive numbers. They can be divided in 3 groups as
$G_1:r,r+3,r+6,\dots \dots r+3n-3G_2:r+1,r+4,r+7,\dots \dots r+3n-2G_3:r+2,r+5,r+8,\dots \dots r+3n-1$
Sum of three numbers is divisible by 3 if all 3 are from $G_1$ or $G_2$ or $G_3$, or exactly one is from each of $G_1$, $G_2$ and $G_3$.
Probability$=\frac{{}^n{C}_3\times 3{+}^n{C}_1{\times }^n{C}_1{\times }^n{C}_1}{3^n{C}_3}$
$=\frac{3n^2-3n+2}{(3n-1)(3n-2)}$