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

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===Solution===
 
===Solution===
 
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>{\frac{2}{3}}^2=\frac{4}{9}</math> of having an odd product. Thus, there is a <math>1-\frac{4}{9}=\frac{5}{9}</math> probability of having an even product. We get our answer to be <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
 
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>{\frac{2}{3}}^2=\frac{4}{9}</math> of having an odd product. Thus, there is a <math>1-\frac{4}{9}=\frac{5}{9}</math> probability of having an even product. We get our answer to be <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
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===Solution 2===
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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 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 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 3, 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  <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>.
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==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=6|num-a=8}}
 
{{AMC8 box|year=2015|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:56, 26 November 2015

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?

$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$

Solution

We can instead find 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 ${\frac{2}{3}}^2=\frac{4}{9}$ of having an odd product. Thus, there is a $1-\frac{4}{9}=\frac{5}{9}$ probability of having an even product. We get our answer to be $\boxed{\textbf{(E) }\frac{5}{9}}$.


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 $1$, 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 $\boxed{\textbf{(E) }\frac{5}{9}}$.

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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