Difference between revisions of "1991 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | <math>\ | + | Solution by e_power_pi_times_i |
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| + | Because the odds against <math>X</math> are <math>3:1</math>, the chance of <math>X</math> losing is <math>\dfrac{3}{4}</math>. Since the chance of <math>X</math> losing is the same as the chance of <math>Y</math> and <math>Z</math> winning, and since the odds against <math>Y</math> are <math>2:3</math>, <math>Y</math> wins with a probability of <math>\dfrac{3}{5}</math>. Then the chance of <math>Z</math> winning is <math>\dfrac{3}{4} - \dfrac{3}{5} = \dfrac{3}{20}</math>. Therefore the odds against <math>Z</math> are <math>\boxed{\textbf{(D) } 17:3}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 12:05, 13 December 2016
Problem
Horses 
and 
are entered in a three-horse race in which ties are not possible. The odds against 
winning are 
and the odds against 
winning are 
, what are the odds against winning? (By "odds against 
winning are 
" we mean the probability of winning the race is 
.)
Solution
Solution by e_power_pi_times_i
Because the odds against are , the chance of losing is 
. Since the chance of losing is the same as the chance of and winning, and since the odds against are , wins with a probability of 
. Then the chance of winning is 
. Therefore the odds against are 
.
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
