1991 AMC 12 Problem 13

Attempt Problem 13 of the 1991 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

13.

Horses

X,

Y,

and

Z

are entered in a three-horse race in which ties are not possible. If the odds against

X

winning are

3

-to-

1

and the odds against

Y

winning are

2

-to-

3,

what are the odds against

Z

winning? (By “odds against

H

winning are

p

-to-

q

” we mean that the probability of

H

winning the race is

\frac{q}{p+q}.

)

3

-to-

20

5

-to-

6

8

-to-

5

17

-to-

3

20

-to-

3

Convert each set of odds to a probability

The three winning probabilities sum to

1

The given probabilities are

P(X)=\frac14

and

P(Y)=\frac35.

Therefore

P(Z)=1-\frac14-\frac35=\frac3{20}.

The probability that

Z

loses is

\frac{17}{20},

so the odds against

Z

are

17

-to-

3.

Thus the correct answer is D.

Problem 13 in other years