Question

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?

$19$

$21$

$46$

$69$

more than $69$

Answer: B

Concepts:combinations Diophantine equations

Difficulty rating: 2190

Small Hint:

Let the four color counts be $r,w,b,g$ and equate the combination counts for the four events

Big Hint:

Successive ratios determine $w,b,g$ in terms of $r$ ; then find the smallest $r$ making all three integers

Solution:

Equal probabilities have the same common denominator, so their favorable selection counts satisfy $\binom r4=w\binom r3$ $=wb\binom r2=wbgr.$ Successive ratios give $w=\frac{r-3}{4},$ $b=\frac{r-2}{3},$ and $g=\frac{r-1}{2}.$ The least $r\ge4$ making all three positive integers is $r=11.$ Then $(w,b,g)=(2,3,5),$ for $11+2+3+5=21$ marbles. Thus the correct answer is B.

1996 AMC 12 Problem 26

Answer: B

Difficulty rating: 2190

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years