A bag contains some marbles, each of which is one of four colors (red, white, blue, and green). There is at least one of each color. The composition of the bag is such that if we take four marbles out at random (without replacement), each of the following is equally likely:
(1) one marble of each color is chosen,
(2) one white, one blue, and two reds are chosen,
(3) one blue and three reds are chosen,
(4) four reds are chosen.
What is the smallest possible number of marbles in the bag?
+6248 \(\text{In general the probability of an assortment of 4 marbles is given by}\\ p=\dfrac{\dbinom{n_r}{k_r}\dbinom{n_w}{k_w}\dbinom{n_b}{k_b}\dbinom{n_g}{k_g}}{\dbinom N 4}\\ \text{where $n_*$ is the number of * colored marbles in the bag, and $k_*$ is how many are picked of color *}\\~\\ \text{Setting all these equal, and noting that $\dbinom n 1 = n$ we get}\\~\\ n_r n_w n_b n_g = n_w n_b \dbinom{n_r}{2} = n_b \dbinom{n_r}{3} = \dbinom{n_r}{4}\)
\(\text{This can be distilled down into}\\ n_b = \dfrac{\dbinom{n_r}{4}}{\dbinom{n_r}{3}}\\ n_w = \dfrac{\dbinom{n_r}{3}}{\dbinom{n_r}{2}}\\ n_g = \dfrac{\dbinom{n_r}{2}}{n_r}\\ \text{and all of these must be positive integer values}\)
\(\text{The first value of $n_r$ for which this is true is $n_r=11$}\\ n_r=11,~n_w=3, n_b=2, n_g = 5\\~\\ \text{This is a total of 21 marbles in the bag}\)
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