# Red And Green Balls in Red And Green Boxes

### Problem

### Solution

Let $g$ stand for the number of green balls in the red box. Then, since the green box ought to have at least one green ball, we have the following distribution of colors:

$\begin{array}{cc} \text{Red box (}5\text{ balls)} & \text{Green box (}9\text{ balls)}\\ g\text{ green} & 8-g\text{ green}\\ 5-g\text{ red} & g+1\text{ red} \end{array}$

Therefore, the number of green balls in the red box plus the number of red balls in the green box equals $g+(g+1)=2g+1.$

Note that, since $0\le g\le 5,$ $1\le 2g+1\le 11.$ The only non-prime odd numbers in this range are $1$ and $9;$ for these, $g=0$ or $g=4.$ The probability of drawing all five red balls or just one red ball is given by

$\displaystyle \frac{\displaystyle {6\choose 5}+{8\choose 4}{6\choose 1}}{\displaystyle {14\choose 5}}=\frac{6+420}{2002}=\frac{213}{1001}.$

### Acknowledgment

This is a chapter in R. Honsberger's Mathematical Morsels0883853035 (MAA, 1978). Honsberger refers to Problem E1400 from the AMM (1960, p 698) by C. W. Trigg.

|Contact| |Front page| |Contents| |Probability|

Copyright © 1996-2018 Alexander Bogomolny66868089