Question

Out of 3n consecutive integers, three are selected at random. Find the chance chat their sum, is divisible by 3.

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

Answer 1

Answer: (D)

$\frac{3n^2-3n+2}{(3n-1)(3n-2)}$

Let the sequence of numbers start with the integer m so that the 3n consecutive integers are $m,m+1,m+2,m+3,\cdots ,m+3n-1.$
Now they can be classified as $m,m+3,m+6,...,m+3n-2,m+2,m+5,m+8,....,m+3n-1$.
The sum of the three numbers shall be divisible by 3 if either all the three numbers are from the same row or all the three numbers are from the same row or all the three numbers are from different rows.
The number of ways that the three numbers of ways that the numbers are from different rows is $n\times n\times n=n^3$
since a number can be selected from each row in n ways.Hence the favourable no.of ways $M={3.}^n{C}_3+n^3$
And the total number of ways $N{=}^{3n}{C}_3$
$\therefore$ The required probability $=\frac{M}{N}=\frac{{3.}^n{C}_3+n^3}{{}^{3n}{C}_3}=\frac{3n^2-3n+2}{(3n-1)(3n-2)}$