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Q.
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
Probability
Solution:
Let us divide given $3\, n$ 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 $= 3 \times \,{}^{n}C_{3} + n^{3}$
$= \frac{3.n\left(n -1\right)\left(n -2\right)}{3.2.1} +n^{3} = \frac{3n^{3} -3n^{2} +2n}{2}$
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)}{3. 2.1} = \frac{n\left(3n -1\right) 3n-2}{2}$
$\therefore $ Required Probability$ = \frac{\text{Favourable ways}}{\text{Total ways}}$
$= \frac{3n^{3} -3n^{2} +2n}{n\left(3n -1\right) \left(3n -2\right)} = \frac{3n^{2} -3n +2}{\left(3n -1\right) \left(3n -2\right)}$