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

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===Solution 2===
 
===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>.
+
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>.
  
 
==See Also==
 
==See Also==

Revision as of 11:56, 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 (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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