Q.6.61 - A First Course in Probability

Question

Consider an urn containing n balls numbered $1,.....,n$ and suppose that k of them are randomly withdrawn. Let $X_i$ equal $1$ if ball number $i$ is removed and let $X_i$ be $0$ otherwise. Show that $X_1,......X_n$ are exchangeable .

Step-by-Step Solution

Answer

The probability does not depend on permutation. Hence the variables are exchangeable.

Step 1 : Variable :

The numerical value or quantity expressed by an alphabetic letter. 

Step 2 : Explanation :

$X_i=1,$ if ball $i$ is removed.

And

$X_i=0,$ if ball is $i$ not removed.

Only k out of these variables assume value $X_i=1$.

And other assume value $X_i=0$.

Thus, for choosing more or less than k ones in n integers $i_1,.....,i_{n,}$

$P({\cap }_i{X}_i=i_i)=0$

Since this event is an impossible event in every permutation.

If we have exactly k ones in n integers $i_1,......i_n,$

$P({\cap }_i{X}_i=i_i)=\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}$

All possible configurations of k chosen balls are equally likely due to symmetry.

And there exists $\left(\begin{array}{c}n\\ k\end{array}\right)$ of these combinations.

Thus,

It has been proved that the probability does not depend on permutation.

Hence, the variables are exchangeable.