In a bucket, there are five different colors of balls, two of each color, making 10 in total. If you pick three balls at random without replacement, what is the probability that you pick a different colored ball each time? What is the probability that you only pick two different colored balls?
I just made up a simple random scenario that demonstrates the principle/type of problem that I am trying to figure out. With the skills I acquired in pre-calculus, I would be able to solve this if it was with replacement, not without replacement. Also having several unique colors, but identical balls within each color group adds an extra layer of complexity that confuses me. What is the process to solve the problem? And, if I was to increase the number of colors, number of balls per color, or the number of times a ball is picked, could the same formula/process still be applied?
Best Answer
If you're picking the balls without replacement, then you can treat this problem with combinations (since it doesn't matter in which order you pick the balls).
There are $_{10}C_3$ ways to pick three balls out.
To pick three different colors, choose the colors ($_5C_3$) and then choose which ball of each color ($8$). This gives probability $8\cdot (_5C_3) / (_{10}C_3)$.
To pick two different colors, pick the color with two balls ($5$) and pick the other ball ($8$). So, probability $5 \cdot 8 / (_{10}C_3).$
This method works for any number of colors and any number of balls per color, as long as the number of balls per color is the same. If you allow for different numbers of balls of particular colors then you'll need to treat those cases differently.