Each of two boxes contains three chips numbered $1$, $2$, $3$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? 
$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$
Solution 1.1
(This solution is similar to Solution 2.)
Let's make this a problem with boxes. 
In total, there are 9 products derived from these numbers (because 3 numbers per box). This would be our denominator. 
Every time a spinner lands on 2, we get an even product. These are $(1,2)$, $(2,1)$, $(2,2)$, $(2,3)$, and finally $(3,2)$. 
Going back, we see there are $3\cdot3=9$ possible combinations, and we have $5$ evens, the final answer is  
$\boxed{\textbf{(E) }\frac{5}{9}}$.

Solution 1
You can also make this problem into a spinner problem. You have the first spinner with $3$ equally divided 
sections: $1, 2$, and $3$. You make a second spinner that is identical to the first, with $3$ equal sections of 
$1$,$2$, and $3$. If the first spinner lands on $1$, it must land on two for the result to be even. You write down the first 
combination of numbers: $(1,2)$. Next, if the spinner lands on $2$, it can land on any number on the second 
spinner. We now have the combinations of $(1,2) ,(2,1), (2,2),$ and $(2,3)$. Finally, if the first spinner ends on $3$, we 
have $(3,2).$ Since there are $3\cdot3=9$ possible combinations, and we have $5$ evens, the final answer is  
$\boxed{\textbf{(E) }\frac{5}{9}}$.

Solution 3
We can also list out the numbers. Box A has chips $1$, $2$, and $3$, and Box B also has chips $1$, $2$, and $3$. Chip $1$ (from Box A) 
could be with 3 partners from Box B. This is also the same for chips $2$ and $3$ from Box A. $3+3+3=9$ total sums. Chip $1$ could be 
multiplied with 2 other chips to make an even product, just like chip $3$. Chip $2$ can only multiply with 1 chip.  $2+2+1=5$. The answer is $\boxed{\textbf{(E) }\frac{5}{9}}$.

Solution 4
Here is another way:
Let's start by finding the denominator: Total choices.
There are $3$ chips we can choose from in the 1st box, and $3$ chips we can choose from in the 2nd box. We do $3*3$, and get $9$.
Now - to find the numerator: Desired choices.
To get an even number, we need to pick 2 from at least one of the boxes. There are $2$ choices as to finding which box we will draw the 2 from. Then, we have $3$ choices from the other box to pick any of the other chips, $1, 2,$ and $3$. 
$\frac{3 \cdot2}{9} = \frac{6}{9}$
However, we are over counting the $(2,2)$ configuration twice, and so, we subtract that one configuration from our total.
$\frac{6}{9} - \frac{1}{9}$.
Thus, our answer is $\boxed{\textbf{(E) }\frac{5}{9}}$.
~ del-math.

Solution 5
This might take longer to solve, but you definitely will have the right answer.
You first list out all the possible combinations regardless if the product is even or odd.
$1 \cdot 1 = 1.$
$1 \cdot 2 = 2.$
$1 \cdot 3 = 3.$
$2 \cdot 1 = 2.$
$2 \cdot 2 = 4.$
$2 \cdot 3 = 6.$
$3 \cdot 1 = 3.$
$3 \cdot 2 = 6.$
$3 \cdot 3 = 9.$   
There are 9 possible combinations, and $5$ of the combinations’ products are even.  So, our answer is $\boxed{\textbf{(E) }\frac{5}{9}}$.
~MiracleMaths