# 2015 AMC 8 Problems/Problem 7

## Problem

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?

$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$

## Solutions

### 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

## Video Solution (HOW TO THINK CRITICALLY!!!)

~Education, the Study of Everything

~savannahsolver