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=== | ||
| + | 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 11:56, 26 November 2015
Each of two boxes contains three chips numbered 
, 
, 
. 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?
Solution
We can instead find the probability that their product is odd, and subtract this from . In order to get an odd product, we have to draw an odd number from each box. We have a 
probability of drawing an odd number from one box, so there is a 
of having an odd product. Thus, there is a 
probability of having an even product. We get our answer to be 
.
Solution 2
You can also make this problem into a spinner problem. You have the first spinner with equally divided sections, 
and 
You make a second spinner that is identical to the first, with equal sections of ,, and . If the first spinner lands on , to be even, it must land on two. You write down the first combination of numbers 
. Next, if the spinner lands on , it can land on any number on the second spinner. We now have the combinations of 
. Finally, if the first spinner ends on 3, we have 
Since there are 
possible combinations, and we have 
evens, the final answer is .
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. 
