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?

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 $3 \cdot {20 \choose 2}$
