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Difference between revisions of "1991 AHSME Problems/Problem 13"

(Created page with "== Problem == Horses <math>X,Y</math> and <math>Z</math> are entered in a three-horse race in which ties are not possible. The odds against <math>X</math> winning are <math>3:1<...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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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 $X,Y$ and $Z$ are entered in a three-horse race in which ties are not possible. The odds against $X$ winning are $3:1$ and the odds against $Y$ winning are $2:3$, what are the odds against $Z$ winning? (By "odds against $H$ winning are $p:q$" we mean the probability of $H$ winning the race is $\frac{q}{p+q}$.)

$\text{(A) } 3:20\quad \text{(B) } 5:6\quad \text{(C) } 8:5\quad \text{(D) } 17:3\quad \text{(E) } 20:3$

Solution

Solution by e_power_pi_times_i


Because the odds against $X$ are $3:1$, the chance of $X$ losing is $\dfrac{3}{4}$. Since the chance of $X$ losing is the same as the chance of $Y$ and $Z$ winning, and since the odds against $Y$ are $2:3$, $Y$ wins with a probability of $\dfrac{3}{5}$. Then the chance of $Z$ winning is $\dfrac{3}{4} - \dfrac{3}{5} = \dfrac{3}{20}$. Therefore the odds against $Z$ are $\boxed{\textbf{(D) } 17:3}$.

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

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