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

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==Problem==
 
==Problem==
  
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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?
+
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>
  
 
==Solutions==
 
==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 <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 1===
 
===Solution 1===
−
We can instead calculate 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> probability 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 2===
 
 
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===
 
===Solution 3===
−
We can also list out the numbers. Hat A has chips <math>1</math>, <math>2</math>, and <math>3</math>, and Hat B also has chips <math>1</math>, <math>2</math>, and <math>3</math>. Chip <math>1</math>(from Hat A)  
+
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 Hat B. This is also the same for chips <math>2</math> and <math>3</math> from Hat A. <math>3+3+3=9</math> total sums. Chip <math>1</math> could be  
+
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>.
 
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>.
  
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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>.
 
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.
 
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, 3</math>.  
+
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>
 
<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.
+
 
 +
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>.
 
<math>\frac{6}{9} - \frac{1}{9}</math>.
 
Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 
Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
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This might take longer to solve, but you definitely will have the right answer.
 
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.
 
You first list out all the possible combinations regardless if the product is even or odd.
−
1 x 1 = 1
+
<math>1 \cdot 1 = 1.</math>     
−
1 x 2 = 2
+
<math>1 \cdot 2 = 2.</math>     
−
1 x 3 = 3
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<math>1 \cdot 3 = 3.</math>   
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2 x 1 = 2
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<math>2 \cdot 1 = 2.</math> 
−
2 x 2 = 4
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<math>2 \cdot 2 = 4.</math>   
−
2 x 3 = 6
+
<math>2 \cdot 3 = 6.</math> 
−
3 x 1 = 3
+
<math>3 \cdot 1 = 3.</math>
−
3 x 2 = 6
+
<math>3 \cdot 2 = 6.</math> 
−
3 x 3 = 9
+
<math>3 \cdot 3 = 9.</math> 
−
There are 9 possible combinations, and 5 of the combinations’s products are even.  So, our answer is our answer is <math>\boxed{\textbf{(E) }\frac{5}{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
+
 
 +
~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

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