Difference between revisions of "1998 AJHSME Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | The different possible values | + | The different possible (and equally likely) values Tamika gets are: |
<math>8+9=17</math> | <math>8+9=17</math> | ||
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<math>9+10=19</math> | <math>9+10=19</math> | ||
− | The different possible values | + | The different possible (and equally likely) values Carlos gets are: |
<math>3\times5=15</math> | <math>3\times5=15</math> | ||
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The probability that if Tamika had the sum <math>17</math> her sum would be greater than Carlos's set is <math>\frac{1}{3}</math>, because <math>17</math> is only greater than <math>15</math> | The probability that if Tamika had the sum <math>17</math> her sum would be greater than Carlos's set is <math>\frac{1}{3}</math>, because <math>17</math> is only greater than <math>15</math> | ||
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The probability that if Tamika had the sum <math>18</math> her sum would be greater than Carlos's set is <math>\frac{1}{3}</math>, because <math>18</math> is only greater than <math>15</math> | The probability that if Tamika had the sum <math>18</math> her sum would be greater than Carlos's set is <math>\frac{1}{3}</math>, because <math>18</math> is only greater than <math>15</math> | ||
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The probability that if Tamika had the sum <math>19</math> her sum would be greater than Carlos's set is <math>\frac{2}{3}</math>, because <math>19</math> is greater than both <math>15</math> and <math>18</math> | The probability that if Tamika had the sum <math>19</math> her sum would be greater than Carlos's set is <math>\frac{2}{3}</math>, because <math>19</math> is greater than both <math>15</math> and <math>18</math> | ||
Revision as of 11:03, 31 July 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 (and equally likely) values Tamika gets are:
The different possible (and equally likely) values Carlos gets are:
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 |