# Finding the distribution of random variables

• LokLe
Your distribution of Y adds up to more than 1, so we can see it must be wrong. Furthermore, it does not make sense that the probabilities of the X values are equal. Can the maximum being 1 (only two ways to get that) really be as likely as the maximum being 9 (more than two ways to get that)?Your distribution of Y adds up to more than 1, so we can see it must be wrong. Furthermore, it does not make sense that the probabilities of the X values are equal. Can the maximum being 1 (only two ways to get that) really be as likely as the maximum being 9 (more than two ways to get that)?f

#### LokLe

Homework Statement
In a box of ten tickets numbered 0, 1, 2,...,9. You pick at random 2 tickets from this box. Let
X is the larger number in these two tickets, and Y be the smaller one.
(a) Find the distributions of X, Y and their joint distribution.
(b) Find the distribution of Z=X-Y.
(c) Compute E(X), Var(X), E(Z) and Var(Z).
Relevant Equations
None
Hi. I have found the answer to a and c (I don't know whether it is correct) but I do not know what I should find in question b.

Is my method correct and how should I solve part b?

I made a 10x10 table and counted the occurrence of each max and min. There is a simple pattern.
I don't get the same answers in (a). Can you explain what your logic is? I have trouble reading your work, but it looks like your probabilities add up to more than 1. [EDIT: It looks like I agree with the numbers at the top but not with the numbers in the box.]

Last edited:
anuttarasammyak
I also made a table with number Z in the cells.

Last edited:
WWGD and FactChecker
I made a 10x10 table and counted the occurrence of each max and min. There is a simple pattern.

I tried to correct the mistakes by making a 10x10 table.

We have the same chance of getting a number from 0 to 9, so the distribution of (X=x) should be 1/10.
Since we have already picked a number for X, (Y=y) should be 1/(10-1) = 1/9?

I also made a table with number Z in the cells.View attachment 298596
I understand part b now. Thank you!

I tried to correct the mistakes by making a 10x10 table.

We have the same chance of getting a number from 0 to 9, so the distribution of (X=x) should be 1/10.
Since we have already picked a number for X, (Y=y) should be 1/(10-1) = 1/9?
You pick up two tickets AT A TIME. Then you read them to call larger one Y and smaller one X.
All the possible cells in the table have an equal probability.

You pick up two tickets AT A TIME. Then you read them to call larger one Y and smaller one X.
All the possible cells in the table have an equal probability.
Oh so X=x and Y=y have an equal distribution. I will correct it. Thank you.

Oh so X=x and Y=y have an equal distribution. I will correct it. Thank you.
I would say all the cells (x,y) x<y are equal in probability or in distribution.

LokLe
Then you read them to call larger one Y and smaller one X.
You got that backwards.

anuttarasammyak