Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by $$3$$ is :

Solution :
Let $${E_3} = $$  the event of the sum being $$3.$$ Similarly, $${E_6},\,{E_9},\,{E_{12}},\,{E_{15}},\,{E_{18}},\,{E_{21}}.n\left( S \right) = {}^{12}{C_2}$$
$$\eqalign{ & n\left( {{E_3}} \right) = 1,\,n\left( {{E_6}} \right) = 2,\,n\left( {{E_9}} \right) = 4,\,n\left( {{E_{12}}} \right) = 5, \cr & n\left( {{E_{15}}} \right) = 5,\,n\left( {{E_{18}}} \right) = 3,n\left( {{E_{21}}} \right) = 2 \cr & \therefore \,P\left( E \right) = P\left( {{E_3}} \right) + ...... + P\left( {{E_{21}}} \right) \cr & = \frac{{1 + 2 + 4 + 5 + 5 + 3 + 2}}{{{}^{12}{C_2}}} \cr & = \frac{{22 \times 2}}{{12 \times 11}} \cr & = \frac{1}{3} \cr} $$
Alternatively : The sum of the numbers for every selection is divisible by $$3$$ or leaves the remainder $$1$$ or leaves the remainder $$2.$$ These are equally probable. So, the required probability $$ = \frac{1}{3},$$  because the sum of the three probabilities is $$1.$$