1991 AHSME Problems/Problem 13

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 $\boxed{\textbf{(D) } 17:3}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1991 AHSME Problems/Problem 13

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