Probability that sum is odd but not divisible by $3$ Out of $20$ consecutive natural numbers two are chosen randomly.Find  Probability that sum is odd but not divisible by $3$.
We have denominator as $\binom{20}{2}$.
Now any number will be of the form $3k$, $3k+1$ or $3k-1$.
Since sum of $3p+1$ and $3q-1$ is divisible by $3$ our chosen two numbers must fall in the following two cases:
Case $1.$ one number is of form $3k$ and another $3p+1$
So select any multiple of $3$ from $20$ numbers and select any number which leaves remainder $1$ from among $20$ which makes sum not divisible by 3.
Case $2.$ one number is of form $3k$ and another $3q-1$
So select any multiple of $3$ from $20$ numbers and select any number which leaves remainder $2$ from among $20$ which makes sum not divisible by 3.
But now how to choose making their sum is odd?

Note: $\Bbb N = 6\Bbb N+3\{0,1\}+\{0,1,2\}$ and $\Bbb N\times\{0,1\}\times\{0,1,2\}\mapsto \Bbb N$ is a bijection.  
Ie: we can uniquely identify every natural number by the form $6a_1+3a_2+a_3$ where $a_1\in\Bbb N,$ $a_2\in\{0,1\}$, and $a_3\in\{0,1,2\}$. Eg: six consecutive natural numbers starting at $6k$ are: $$\{6k, 6k+1,6k+2,6k+3, 6k+3+1, 6k+3+2\}$$
The sum of two numbers of such form, $(6a_1+3a_2+a_3)$ and $(6b_1+3b_2+b_3)$, will be odd but not divisible by three if:


*

*$a_2+b_2=1$ and $a_3+b_3 \in\{2,4\}$, that is $(a_2,b_2)\in\{(0,1),(1,0)\}\text{ and }\\(a_3,b_3)\in\{(0,2),(1,1),(2,0),(2,2)\}$

*$a_2+b_2\in\{0,2\}$ and $a_3+b_3 = 1$, that is $(a_2,b_2)\in\{(0,0),(1,1)\}\text{ and }\\(a_3,b_3)\in\{(0,1),(1,0)\}$

