1996 AMC 12 Problem 26
Attempt Problem 26 of the 1996 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 1996 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
26.
An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
more than
Answer: B
Small hint:
Let the four color counts be and equate the combination counts for the four events
Big hint:
Successive ratios determine in terms of ; then find the smallest making all three integers
Solution:
Equal probabilities have the same common denominator, so their favorable selection counts satisfy Successive ratios give and The least making all three positive integers is Then for marbles. Thus the correct answer is B.
Problem 26 in other years
- 1950 AMC 12
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