Difference between revisions of "1998 AJHSME Problems/Problem 19"
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<math>\text{(A)}\ \dfrac{4}{9} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{3}</math> | <math>\text{(A)}\ \dfrac{4}{9} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{3}</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 19:53, 26 October 2016
Problem
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.