Difference between revisions of "1998 AJHSME Problems/Problem 19"
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<math>8+9=17</math> | <math>8+9=17</math> | ||
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<math>8+10=18</math> | <math>8+10=18</math> | ||
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<math>9+10=19</math> | <math>9+10=19</math> | ||
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<math>3\times5=15</math> | <math>3\times5=15</math> | ||
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<math>3\times6=18</math> | <math>3\times6=18</math> | ||
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<math>5\times6=30</math> | <math>5\times6=30</math> | ||
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<math>\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}</math> | <math>\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}</math> | ||
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== See also == | == See also == |
Revision as of 13:07, 10 June 2011
Problem 19
Tamika selects two different numbers at random from the set and adds them. Carlos takes two different numbers at random from the set and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?
Solution
The different possible values of Tamika's set leaves:
The different possible values of Carlos's set leaves:
The probability that if Tamika had the sum her sum would be greater than Carlos's set is , because is only greater than The probability that if Tamika had the sum her sum would be greater than Carlos's set is , because is only greater than The probability that if Tamika had the sum her sum would be greater than Carlos's set is , because is greater than both and
Each sum has a possibility of being chosen, so we have
See also
1998 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 |