Question: " You have $3$ green balls, $1$ purple ball and $4$ red balls, all in a bag. What is the probability that when you (without replacement) draw $4$ balls, $3$ of them will be green and none of them will be red."
Basically, what is the probability that you draw all the green and purple balls?
Also, is this a permutation or combination question?
What I have tried: $\frac{4P4}{8P4} = \frac{1}{70}$.
Not sure if that is correct or not because it doesn't feel right. NOTE: I get the same answer if I use choose instead.
Best Answer
Your answer is correct. I'll show my attempt using a slightly different method. First,
let $g = green, \; p = purple, \; and \; r = red$
Then, we can realize that the ways of drawing all of the green balls, and then all of the purple balls is:
$1)\; G \; G \; G \; P$
$2)\; G \; G \; P \; G$
$3)\; G \; P \; G \; G$
$4)\; P \; G \; G \; G$
If we then consider the first possibility, $ G \; G \; G \; P$:
$= \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} \times \frac{1}{5}$
$= \frac{1}{280}$
Because the possibility of each of the four sequences is equal in this case, the total probability of getting all the green balls and all of the purple balls is
$4 \times \frac{1}{280}$
$=\frac{1}{70}$
Which is the answer you got :)