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?
 A: Because with $20$ numbers you do not have the same number of numbers of the form $3k$, $3k+1$, and $3k+2$, I would recommend considering three separate cases, namely where the series starts with a number of the form $3k$, where it starts with $3k+1$, and where it starts with $3k+2$. For each, figure out how many pairs there would be of the desired property (because of the asymmetry, probably you don't get the same number for each), add them all up, and divide by $3 \cdot {20 \choose 2}$
Just to show how to do this for one case, consider where the first number of the series is of the form $3k$. Now, to add a second number and get an odd sum, we need to either add the second number, or the fourth number , or ... Of those, the 4th, 10th, and 16th will be of the form $3p$, so we rule those out, leaving $7$ numbers that can be added to the first with the desired property. The same $7$ numbers can be added to the 7th, 13th, and 19th number, giving $28$ pairs. Similarly, to the 3rd, 9th, and 15th we can add all but the 6th, 12th, and 18th, so that is another $21$ pairs. Finally, to the 5th, 11th, and 17th we can add all but the 2nd, 8th, 14th, and 20th, giving another $18$ pairs, for a total of $67$ pairs.
Now do the same analysis for the first number being of the form $3k+1$, and then for $3k+2$. Like I said, you may get a slightly different number of pairs for those, but add them all up, and divide by the total number of possible pairs you can get between all these three different sequences, i.e divide by $3 \cdot {20 \choose 2}$
A: 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 $(a_2,b_2)\in\{(0,0),(1,1)\}\text{ and }\\(a_3,b_3)\in\{(0,1),(1,0)\}$

A: Firstly, you have overlooked two other cases where the criteria holds: 
$$
(3n\pm1) + (3m\pm1)= 3(q)\pm2=3k\mp1
$$
where $k=q+1$, $q=n+m$, and $n,m,q,k \in \mathbb N $.
We call this scenario (1). We can then combine the two cases you identified,
$$ 3n +(3m\pm1)=3k\pm1
$$
where $k=n+m$, and $n,m,,k \in \mathbb N $, as scenario (2).
Now, note that the sum in each case above will only be odd if $3k$ is even, which implies that $k$ must be even. We can deduce from this that in (1), $q$ is odd, so one of $n$ or $m$ must be odd and the other even. In (2), we know that $n$ and $m$ are either both odd or both even. 
In a sequence of $20$ consecutive integers, ten will be odd and ten will be even. There will also be at most $\lceil\frac{20}{3}\rceil=7$ multiples of $3$. In fact, if the sequence begins on a number $3n$ or $3n-1$ there will be seven multiples of 3, but only six if it begins on a number $3n+1$. We will show that in both cases, the probability of sum $3k\pm1$ being odd is the same. 
If there are seven multiples of $3$ in the sequence, either four are odd or four are even. Assume w.l.o.g that four are odd. This means there are six odd non-multiples of $3$ and seven even non-multiples of $3$. From (2) [$n$ and $m$ are either both odd or both even] this gives $4\times6=24$ and $3\times7=21$ sums of the required form. 
Now we consider sums of form in (1). Of the six odd non-multiples of 3, three will be of form $3n-1$ and three will be of form $3n+1$. In the even case, there will be either four or three of the form $3n-1$ and correspondingly three or four of the form $3n+1$ (this depends on whether the sequence starts on a multiple of 3). Either way, (1) yields us $3\times4=12$ and $3\times3=9$ sums of the required form. This means in total there are $24+21+12+9=66$ sums.
In the case where the sequence only has six multiples of 3, a similar process delivers $3\times7 + 3\times7 + 4\times3 + 4\times3 =66$ sums, so it does not matter what number the sequence begins on. 
All that is left is to find the probability, which is $\frac{66}{{20}\choose{2}}\approx 0.347$.          
