# Thread: A mean that doesn't exist

1. ## A mean that doesn't exist

Hello everyone, I've came across this problem and I need to see if my approach is right...

An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let $X$ denote the selection number of the 1st Black ball chosen.

(a)Find the pdf of $X$.

(b)Show that the $E(X)$ doesn't exist for $X$.

for (a), I thought that if I came accross a black ball I'll replace it so
$P(X=1)=\frac{1}{2},\ P(X=2)=(\frac{1}{2})(\frac{1}{2})$ and for any $k,\ P(X=k)={(\frac{1}{2})}^k$.

As for (b), $E(X)=\sum_k k{(\frac{1}{2})}^k$ which is 2!!
How could this be?!

Any help is very much appreciated!

2. Originally Posted by rebghb
Hello everyone, I've came across this problem and I need to see if my approach is right...

An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let $X$ denote the selection number of the 1st Black ball chosen.

(a)Find the pdf of $X$.

(b)Show that the $E(X)$ doesn't exist for $X$.

for (a), I thought that if I came accross a black ball I'll replace it so
$P(X=1)=\frac{1}{2},\ P(X=2)=(\frac{1}{2})(\frac{1}{2})$ and for any $k,\ P(X=k)={(\frac{1}{2})}^k$.

As for (b), $E(X)=\sum_k k{(\frac{1}{2})}^k$ which is 2!!
How could this be?!

Any help is very much appreciated!
The probability that the second ball is B given the first is W is $1/3$ not $1/4$

In general the probability that the first B ball is seen on the $N$-th selection is:

$P(N)=\frac{1}{N(N-1)}$

CB

3. Okay but noone implied that there is a conditional probability... The probavility of getting a black ball is the same as that of getting a white ball which is 1/2...
What about the 1st selection, we would have a pole!!

4. Originally Posted by rebghb
Okay but noone implied that there is a conditional probability... The probavility of getting a black ball is the same as that of getting a white ball which is 1/2...
No it is not, you are adding white balls to the urn until you get a black ball, so the probabilities change for every draw. For the N-th draw to be the first B requires that the previous N-1 were all W.

What about the 1st selection, we would have a pole!!
it is only valid for N>1, what I wrote was the solution for the recurrence implied in the problem with initial condition p(1)=1/2 (or rather the initial state being (1,1)).

If the first draw is W the urn contains 2W and 1B for the second draw so the probability of B on the second draw given that the first was W is 1/3. So the probability that the first B occurs on the 2-nd draw is (1/2)(1/3)

CB

5. I would like to start by thanking you (oh I will press the "thanks" button), BUT!...

This is my post:
An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let denote the selection number of the 1st Black ball chosen.
This is how you interpreted the problem:
you are adding white balls to the urn until you get a black ball, so the probabilities change for every draw. For the N-th draw to be the first B requires that the previous N-1 were all W.
I don't get how you translated what I said (which was completely vague) into something interesting... Thanks though and you can consider it a closed thread...