Probability that the sum of two integers is even between 20 and 40 Question: What is the probability that the sum of two randomly chosen integers between $20$ and $40$ inclusive is even (the possibility of the two integers being equal is allowed)? 
Comment on the answers (a) $\frac{1}{2}$, (b) $\frac{11}{21}$, and (c) $\frac{11^2}{21^2}$
My attempt so far:
I know there are $11$ even numbers between $20$ and $40$ and $10$ odd numbers between $20$ and $40$. So with even numbers, there is a total of $121$ options and $100$ options between odd numbers. 
https://answers.yahoo.com/question/index?qid=20130608150439AA7kjc3 (This is the link I am working off of to understand the problem).
I understand where the $221$ comes from...where/what is the $441$?
 A: There are for possibilities for the two integers: EE, EO, OE, and OO. For the number of ways of getting each one, we can multiply the number of corresponding parities.


*

*For EE, there are $11$ even numbers, so we have $11\times11$.

*For EO, we have $11$ even numbers and $10$ odd, so that's $11\times10$.

*For OE, we have $10\times11$.

*For OO, we have $10\times10$.
EE represents two even numbers, which sum to an even number. OO represents the sum of two odd numbers, which is also even. So the even possibilities are $$11\times11+10\times10 = 221$$ while the total number of possibilities is $$11\times11+11\times10+10\times11+10\times10 = 441$$ We can also get $441$ by noting that there are $21$ total numbers, so there are $21\times21$ total ways of getting two numbers.
A: You are correct till 221. The logic applied says that if there are x integers which you want to pair(same number twice allowed), you have x options for first no. and x options for second no., giving a total of $x^2$ pairs.
This is how you got $11^2$=121 for even and $10^2$=100 for odd.
likewise, you have 11 even and 10 odd numbers, a total of 21 numbers. So, all possible pairs are $21^2$=441.
So, probability= $221/441$
