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

(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 2

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

Solution 6

We can use complementary counting to solve this problem. If we want to find the number of ways for the product to be odd, then the chip chosen from either box has to be $1$ or $3$ . Since there are 2 ways out of 3 to pick an even chip from a box, the probability of picking an even chip from a box is $\frac{2}{3}$ . Since there are $2$ boxes, we can find that the answer is $\frac{2}{3}\cdot\frac{2}{3}$ = $\boxed{\textbf{(E) }\frac{5}{9}}$ .

~JQTX941016

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
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All AJHSME/AMC 8 Problems and Solutions

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