Drawing balls from a bag (Permutations).

Question

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.

Answer 1

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 :)

Answer 2

In can help to reword the problem so that the objects are distinguishable to see what's going on more clearly. Imagine cards instead of balls. There are eight cards, each with a number from 1 to 8. Four of the cards are red. If you shuffle the cards, what's the probability that all the red cards will be in the bottom half of the pile?

There are $8!$ ways in general to order the cards. There are $(4!)^2$ ways to arrange the cards so that all the red cards are in the bottom half (there are $4!$ ways to arrange the top half, and $4!$ ways to arrange the bottom half). So we have $(4!)^2$ out of $8$!, or $\frac1{70}$.

Another way of seeing it is that there's a $\frac 48$ of the first card being non-red. Then there are $3$ non-red cards left out of $7$, so the probability, conditioned on the first card being non-red, of the second card being non-red is $\frac 37$. And so on. So we have $\frac 48 \cdot \frac 37 \cdot \frac 26 \cdot \frac 15$. That simplifies to $\frac 12 \cdot \frac 37 \cdot \frac 13 \cdot \frac 15$, cancel the 3's to get $\frac 12 \cdot \frac 17 \cdot \frac 15 = \frac 1 {70}$