I have the following Task:
Given is a bag of 8 red, 4 yellow and 3 blue marbles. 4 marbles are randomly taken out of the bag (without putting it back).
What is the probability that among the drawn four marbles all possible colors will appear? So the probability of at least 1 red and at least 1 yellow and at least 1 blue marbles.
I have written a simulation for this specific setup that repeats the experiment 10 million times. The result is that it has about a probability of 0.4219. But I can't figure out how to calculate it in general.
Best Answer
The chance of drawing specifically $RRYB$ is $\frac 8{15}\cdot \frac 7{14}\cdot \frac 4{13} \cdot \frac 3{12}$. There are $\frac {4!}{2!}$ orders for these colors, so the chance of getting two reds, one yellow, and one blue is $\frac 8{15}\cdot \frac 7{14}\cdot \frac 4{13} \cdot \frac 3{12}\cdot \frac {4!}{2!}$
Do the other two choices for the colors you draw and add them up.