Difference between revisions of "2015 AMC 8 Problems/Problem 7"
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<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> | ||
+ | ===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>. | ||
==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 17:39, 25 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 .
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.