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Urn problem

  • #1

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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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The probability that the first ball is labeled [itex]x_1[/itex] for any [itex]x_1[/itex] is 1/n. The probability that the second ball is labeled [itex]x_2[/itex] for any [itex]x_2[/itex] other than [itex]x_1[/tex] is 1/(n-1), etc. Thus the probability [itex]P(X)= P(X_1, X_2, ..., X_n)[/itex][itex]= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n![/itex] 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).
 
  • #3
The probability that the first ball is labeled [itex]x_1[/itex] for any [itex]x_1[/itex] is 1/n. The probability that the second ball is labeled [itex]x_2[/itex] for any [itex]x_2[/itex] other than [itex]x_1[/tex] is 1/(n-1), etc. Thus the probability [itex]P(X)= P(X_1, X_2, ..., X_n)[/itex][itex]= (1/n)(1/(n-1))(1/(n-2))..., (1/2)(1/1)= 1/n![/itex] 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?
 

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