Q:

Probability: Independent or Dependent for uniform random element sets

Question:  You choose a uniformly random element, say $x$, from the set $\{1,2,3\}$ and you choose a uniformly random element, say $y$, from the same set $\{1,2,3\}$. ($x$ and $y$ are chosen independently of each other.) Define the events
$A$ = “$x$ is odd",
$B$ = “$x + y$ is odd".
Are these independent or not? Show proof.

Attempt:
I got $\Pr(A) = 2/3$ (2 odd numbers $\{1,3\}$ to choose from $\{1,2,3\}$)
$\Pr(B)$ was confusing. I know that the addition of odd numbers is even. So, if $x = 1,3$ and $y=2$ will result in $x+y$ to be even. Then $\Pr($picking $2) = 1/3$ and $\Pr($picking $1,3) = 2/3$
$\Pr(B) = 1 – \left(\left(\frac{1}{3} \times \frac{1}{3}\right) + \left(\frac{2}{3}+\frac{2}{3}\right)\right)$
$\Pr(B) = \frac{4}{9}$
$\Pr(A \cap B) = \frac{2}{3} \times \frac{4}{9}$
So they are independent? The answer is they are not independent.

A:

$P(A) = \text{the probability that the first element chosen is odd} = 2/3$, for the reason you provided.
Next, to compute $P(B)$, note that we can only get an odd sum by having $x$ and $y$ of opposite parity. This can occur by setting $(x, y)$ equal to one of $(1, 2)$, $(3, 2)$, $(2, 1)$, $(2, 3)$. From listing all of the possible ways, we see there are exactly $4$ ways to obtain an odd sum out of $3 \cdot 2 = 6$ ways to choose $x$ and $y$. Thus, $P(B) = 4/6 = 2/3$. 
$P(A \cap B)$ is the probability that the first element chosen is odd and the sum is odd. This can happen only if the second element chosen is even, meaning that the second element is $2$. There are only two ways to achieve this: $(x, y) = (1, 2)$ or $(x, y) = (3, 2)$. Once again, since there are $6$ total ways to choose $x$ and $y$, $P(A \cap B) = 2/6 = 1/3$. 
To check for independence, we have to see whether the equation $P(A) \cdot P(B) = P(A \cap B)$ is satisfied. Since it is not satisfied $(2/3 \cdot 2/3 \neq 1/3$), we conclude that they are not independent.

