Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 1)
 
(38 intermediate revisions by 20 users not shown)
Line 1: Line 1:
−
Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. 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?
+
==Problem==
 +
 
 +
Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. 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?  
  
 
<math>\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}</math>
 
<math>\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}</math>
  
−
===Solution===
+
==Solutions==
−
We can instead find the probability that their product is odd, and subtract this from <math>1</math>. In order to get an odd product, we have to draw an odd number from each box. We have a <math>\frac{2}{3}</math> probability of drawing an odd number from one box, so there is a <math>{\left (\frac{2}{3}\right}^2=\frac{4}{9}</math> of having an odd product. Thus, there is a <math>1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}</math> probability of having an even product.
+
===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 <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(2,3)</math>, and finally <math>(3,2)</math>.
 +
 
 +
Going back, we see there are <math>3\cdot3=9</math> possible combinations, and we have <math>5</math> evens, the final answer is
 +
 
 +
<math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
  
−
===Solution 2===
+
===Solution 1===
 
You can also make this problem into a spinner problem. You have the first spinner with <math>3</math> equally divided  
 
You can also make this problem into a spinner problem. You have the first spinner with <math>3</math> equally divided  
  
−
sections, <math>1, 2</math> and <math>3.</math> You make a second spinner that is identical to the first, with <math>3</math> equal sections of  
+
sections: <math>1, 2</math>, and <math>3</math>. You make a second spinner that is identical to the first, with <math>3</math> equal sections of  
  
−
<math>1</math>,<math>2</math>, and <math>3</math>. If the first spinner lands on <math>1</math>, to be even, it must land on two. You write down the first  
+
<math>1</math>,<math>2</math>, and <math>3</math>. If the first spinner lands on <math>1</math>, it must land on two for the result to be even. You write down the first  
  
−
combination of numbers <math>(1,2)</math>. Next, if the spinner lands on <math>2</math>, it can land on any number on the second  
+
combination of numbers: <math>(1,2)</math>. Next, if the spinner lands on <math>2</math>, it can land on any number on the second  
  
−
spinner. We now have the combinations of <math>(1,2) (2,1) (2,2) (2,3)</math>. Finally, if the first spinner ends on <math>3</math>, we  
+
spinner. We now have the combinations of <math>(1,2) ,(2,1), (2,2),</math> and <math>(2,3)</math>. Finally, if the first spinner ends on <math>3</math>, we  
  
−
have <math>(3,2).</math> Since there are <math>3*3=9</math> possible combinations, and we have <math>5</math> evens, the final answer is   
+
have <math>(3,2).</math> Since there are <math>3\cdot3=9</math> possible combinations, and we have <math>5</math> evens, the final answer is   
  
 
<math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 
<math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 +
 +
===Solution 3===
 +
We can also list out the numbers. Box A has chips <math>1</math>, <math>2</math>, and <math>3</math>, and Box B also has chips <math>1</math>, <math>2</math>, and <math>3</math>. Chip <math>1</math> (from Box A)
 +
 +
could be with 3 partners from Box B. This is also the same for chips <math>2</math> and <math>3</math> from Box A. <math>3+3+3=9</math> total sums. Chip <math>1</math> could be
 +
multiplied with 2 other chips to make an even product, just like chip <math>3</math>. Chip <math>2</math> can only multiply with 1 chip.  <math>2+2+1=5</math>. The answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 +
 +
===Solution 4===
 +
Here is another way:
 +
 +
Let's start by finding the denominator: Total choices.
 +
There are <math>3</math> chips we can choose from in the 1st box, and <math>3</math> chips we can choose from in the 2nd box. We do <math>3*3</math>, and get <math>9</math>.
 +
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 <math>2</math> choices as to finding which box we will draw the 2 from. Then, we have <math>3</math> choices from the other box to pick any of the other chips, <math>1, 2,</math> and <math>3</math>.
 +
 +
<math>\frac{3 \cdot2}{9} = \frac{6}{9}</math>
 +
 +
However, we are over counting the <math>(2,2)</math> configuration twice, and so, we subtract that one configuration from our total.
 +
<math>\frac{6}{9} - \frac{1}{9}</math>.
 +
Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 +
 +
~ 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.
 +
<math>1 \cdot 1 = 1.</math>     
 +
<math>1 \cdot 2 = 2.</math>     
 +
<math>1 \cdot 3 = 3.</math>   
 +
<math>2 \cdot 1 = 2.</math> 
 +
<math>2 \cdot 2 = 4.</math>   
 +
<math>2 \cdot 3 = 6.</math> 
 +
<math>3 \cdot 1 = 3.</math>
 +
<math>3 \cdot 2 = 6.</math> 
 +
<math>3 \cdot 3 = 9.</math> 
 +
There are 9 possible combinations, and <math>5</math> of the combinations’ products are even.  So, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 +
 +
~MiracleMaths
 +
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/oODaaTOHemg
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution==
 +
https://youtu.be/PBBrT9iVCVU
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 15:54, 6 January 2024

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_7&oldid=210074"