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!!!)

https://youtu.be/oODaaTOHemg

~Education, the Study of Everything

Video Solution

https://youtu.be/PBBrT9iVCVU

~savannahsolver

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search