How Does Hypergeometric Distribution Calculate Equal Feathered Arrows Remaining?

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Homework Statement


(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?


Homework 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

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
 
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ivan_x3000 said:

Homework Statement


(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?


Homework 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

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

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?
 
I went with hyper geometric aiming to use up 9 red and 3 green, living 4 of each.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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