# Distribution of X for an Urn with n Balls

• DenisaZsuZsa
In summary, there are n balls in the urn, and the probability of a given ball being extracted is 1/n. The distribution of the extracted balls is P(X=x). The mean of X is (n+1)/2, and the variance is (n-1)2/12.

## Homework Statement

You have an urn that contains n balls labeled with the natural numbers {1,2,3,...,n} and you extract n balls from the urn (with the condition that a ball may not be returned to the urn once drawn). You have to determine the distribution of X=(X1,X2,...,Xn), where Xk is the number of the ball from the k-th extraction.

## Homework Equations

For Z a discrete random variable, that can take the values 1, ..., n

Probability Distribution
p(z)=P(Z=z)

Cumulative Distribution Function
F(z)=P(Z<z)

where z is in {1,...,n}

## The Attempt at a Solution

There are n! possible cases.

Xk (k={1,...,n}) are discrete random variable. Before the k-th extraction in the urn there are n-k+1 balls left with the probability of occurrence = 1/(n-k+1) and Xk are discrete random variable:

1 2 ... n​
X1: ( 1/n 1/n ... 1/n )

considering i1 the number of the ball extracted on the first extraction, we have:

1 ... i1-1 i1 i1+1 .. n​
X2: ( 1/(n-1) ... 1/(n-1) 0 1/(n-1) .. 1/(n-1) )

that means that for i1 the probability of occurrence = 0.
...

before the last extraction there is 1 ball left in urn, that obviously, has the probability of occurrence = 1.

I have no clue what to do next and I need it to demonstrate something for a bigger project. I do not ask you for the solution, I just need some hints.

P.S.: I hope you will understand what I wrote here, I've tried my best. Sorry if you don't, english is not my native language.

The probability that the first ball is labeled $x_1$ for any $x_1$ is 1/n. The probability that the second ball is labeled $x_2$ for any $x_2$ other than $x_1[/tex] is 1/(n-1), etc. Thus the probability [itex]P(X)= P(X_1, X_2, ..., X_n)$$= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n!$ as long as X is a permutation of (1, 2, 3,..., n) and 0 if it is not (that is, if the same integer occurs more than once in the "X"s).

HallsofIvy said:
The probability that the first ball is labeled $x_1$ for any $x_1$ is 1/n. The probability that the second ball is labeled $x_2$ for any $x_2$ other than $x_1[/tex] is 1/(n-1), etc. Thus the probability [itex]P(X)= P(X_1, X_2, ..., X_n)$$= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n!$ as long as X is a permutation of (1, 2, 3,..., n) and 0 if it is not (that is, if the same integer occurs more than once in the "X"s).

That is logic, cause it's: the number of favorable cases/number of posible cases => 1/n!
Also you can't say P(X), you have to say P(X=value) or P(X<value), so the corect way is: P(X=(x1,...xn))=P(X1=x1, ...Xn=xn)

What I do not know is how to put the condition: X is a permutation of (1, 2, ...,n), meaning that I don't not how to express the dependecies between X1 and X2 , between X2 and X3 and so on...

If I didn't have the condition: X is a permutation of (1, 2, ...,n), in other words if they where independent, I could say that E[Xi]=(n+1)/ 2 and Var(Xi)= (n-1)2/12, but they aren't independent.

How can I calculate the mean and Var in this case?

## 1. What is an urn model in mathematics?

An urn model in mathematics is a probabilistic model used to describe a system where objects are selected randomly and without replacement. It is often used to model real-world scenarios such as drawing cards from a deck or selecting balls from an urn.

## 2. What is the distribution of X for an urn with n balls?

The distribution of X for an urn with n balls refers to the probability distribution of the number of successes (or balls of a specific color) in a sample of n draws from the urn. This distribution can be calculated using various methods, such as the hypergeometric distribution or the binomial distribution.

## 3. How is the distribution of X affected by the number of balls in the urn?

The distribution of X is directly affected by the number of balls in the urn. As the number of balls increases, the distribution becomes more spread out and the probability of drawing a specific color decreases. This is because there are more possible outcomes as the sample size increases.

## 4. What is the expected value of X for an urn with n balls?

The expected value of X for an urn with n balls is equal to the number of balls of a specific color in the urn multiplied by the probability of drawing that color in one draw. This can be calculated using the formula E(X) = n * (k/N), where n is the sample size, k is the number of balls of a specific color, and N is the total number of balls in the urn.

## 5. Can the distribution of X be used to make predictions about future draws from the urn?

Yes, the distribution of X can be used to make predictions about future draws from the urn. By analyzing the distribution, we can calculate the probability of drawing a specific color in a given number of draws and make informed predictions about the outcomes of future draws. However, it is important to note that the distribution assumes random and independent draws, so actual results may vary.