Question

Out of $3n$ consecutive natural numbers, $3$ natural numbers are chosen at random without replacement. The probability that the sum of the chosen numbers is divisible by $3$ is

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

Answer 1

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

Let the sequence of $3n$ consecutive integers begin with $m$. Then

$3n$ consecutive integers are
$m,m+1,m+2,....m+(3n-1)$
$3$ integers from $3n$ can be selected in ${{}^{3n}C}_3$ ways
$\therefore$ Total no. of outcomes $={{}^{3n}C}_3$

Now $3n$ integers can be divided into $3$ groups

$G_1:n$ numbers of form $3p$
$G_2:n$ numbers of form $3p+1$
$G_3:n$ numbers of form $3p+2$

The sum of 3 integers chosen from $3n$ integers will be divisible by $3$ if either all the three are from same group or one integer from each group. The no. of ways that the three integers are from same group is ${{}^{n}C}_3+{{}^{n}C}_3+{{}^{n}C}_3{{}^{n}C}_1$ and no. of ways that the integers are from different group is ${{}^{n}C}_1\times {{}^{n}C}_1\times {{}^{n}C}_1$

$\therefore$ favourable cases $=({{}^{n}C}_3+{{}^{n}C}_3+{{}^{n}C}_3)+({{}^{n}C}_1\times {{}^{n}C}_1\times {{}^{n}C}_1)$

$\therefore$ Required probability $=\frac{3.{{}^{n}C}_3+{\left({{}^{n}C}_1\right)}^3}{{{}^{3n}C}_3}=\frac{3n^2-3n+2}{(3n-1)(3n-2)}$