Difference between revisions of "2015 AMC 8 Problems/Problem 7"

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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

We can instead calculate the probability that their product is odd, and subtract this from $1$ . In order to get an odd product, we have to draw an odd number from each box. We have a $\frac{2}{3}$ probability of drawing an odd number from one box, so there is a $\left ( \frac{2}{3} \right )^2=\frac{4}{9}$ probability of having an odd product. Thus, there is a $1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}$ probability of having an even product.

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 $1$ , to be even, it must land on two. 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 $(1,2) ,(2,1), (2,2), (2,3)$ . Finally, if the first spinner ends on $3$ , we

have $(3,2).$ Since there are $3*3=9$ 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. Hat A has chips $1$ , $2$ , and $3$ , and Hat B also has chips $1$ , $2$ , and $3$ . Chip $1$ (from Hat A)

could be with 3 partners from Hat B. This is also the same for chips $2$ and $3$ from Hat 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, $1, 2, 3$ .

$\frac{3 \cdot2}{9} = \frac{6}{9}$ 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 x 1 = 1. 1 x 2 = 2. 1 x 3 = 3. 2 x 1 = 2. 2 x 2 = 4. 2 x 3 = 6. 3 x 1 = 3. 3 x 2 = 6. 3 x 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

@@ Line 59: / Line 59: @@
 There are 9 possible combinations, and 5 of the combinations’ products are even.  So, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
-- MiracleMaths
+—-MiracleMaths
 ==See Also==

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

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