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

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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  
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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  
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<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  
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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  
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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   
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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. 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)  
+
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  
 
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  
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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>.

Revision as of 05:12, 26 January 2023

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 $1$, 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

https://youtu.be/PBBrT9iVCVU

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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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