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 12:07, 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 |