Hyper geometric distrubution?

1. Apr 10, 2014

ivan_x3000

1. The problem statement, all variables and given/known data
(a) At the start of the competition, Shirley has 20 arrows in her quiver (a quiver is a container which holds arrows). 13 of Shirley’s arrows have red feathers, and 7 have green feathers. Arrows are not replaced when they are shot at the target.

(i) At the end of the competition, when Shirley has shot 12 arrows, what is the probability that she has an equal number of red and green feathered arrows remaining in her quiver?

2. Relevant equations
Torn between binomial probability and hyper geometric probability distribution

hyper geometric = (C(r,x)*C((n-r,n-x))/(N,n)
N is the population (12 arrows)
n is the number of events (12 picks)
r is the sample space of the the variable we want to focus on either (either 13 or 7)
x is how many of the focus we want to accumulate so if we want 3/13 reds then the number is 3

3. The attempt at a solution
13/20, 7/20

in order to be equal in 9 red (4/13, 3 green (4/7)

so n(12,12)*(13/20)^9*(7/20)^3
Criteria:independent(assumed), probability is constant(assumed), two possible outcomes1

i was going to use hyper geometric but i have no idea how to make the arrows equal in hyper geomtericc
*Because of random selection this is the right one to use but i have no idea how to make the feathers qual
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 10, 2014

Ray Vickson

Think it through: how many red and green arrows does she start with? How many red and green arrows does she end up with (if the desired event occurs)? So, how many red and green arrows did she use?

3. Apr 10, 2014

ivan_x3000

I went with hyper geometric aiming to use up 9 red and 3 green, living 4 of each.