Question

From the set $S=\left\{1,2,3,4,....,3n\right\}$, three numbers are drawn randomly, the probability that their sum is divisible by $3$, is

• A. $\frac{3n}{\left(3n-1\right)\left(3n-2\right)}$
• B. $\frac{{\left({}^n{C}_1\right)}^3}{\left(3n-1\right)\left(3n-2\right)}$
• C. $\frac{3n^2-3n+2}{9n^2-9n+2}$
• D. $\frac{3n^2+3n+2}{\left(3n-1\right)\left(3n-2\right)}$

Answer 1

Answer: (C)

Let us divide given $3n$ numbers (integers) into three sets as follows
$S_1,=1,4,7,...,3n-2$
$S_2=2,5,8,...,3n-1$
$S_3=3,6,9,...,3n$
Now, sum of selected (drawn) three numbers will be divisible by $3$ if
(i) All the three integers are from same set and their ways are ${}^3{C}_1\times {}^n{C}_3$
(ii) Drawn three numbers are from different set and their ways are${}^n{C}_1\times {}^n{C}_1\times {}^n{C}_1=n^3$
$\therefore$ Favourable number of ways $=3\times {}^n{C}_3+n^3$
$=\frac{3.n\left(n-1\right)\left(n-2\right)}{\mathrm{3.2.1}}+n^3=\frac{3n^3-3n^2+2n}{2}$
Again, the total number of ways of drawing three integers out of $3n$ integers
$={}^{3n}{C}_3=\frac{3n\left(3n-1\right)\left(3n-2\right)}{\mathrm{3.2.1}}=\frac{n\left(3n-1\right)3n-2}{2}$
$\therefore$ Required Probability$=\frac{\text{Favourable ways}}{\text{Total ways}}$
$=\frac{3n^3-3n^2+2n}{n\left(3n-1\right)\left(3n-2\right)}=\frac{3n^2-3n+2}{\left(3n-1\right)\left(3n-2\right)}$