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I am very new in probability and statistics, and I don't know if I can post here in cross-validated this very simple question. However I have this problem:

An urn has $10$ numbered balls such that

  • $4$ of them have the number $0$

  • $2$ of them have the number $1$

  • $4$ of them have the number $2$

Let $X$ a random variable to specify the numeric value a ball $x$ drawn from the set of ten balls, $\Omega$. Determine the probability density function.

I have problems to identify the set of possible outcomes, I considered it as:
$\Omega = \left \{ 1, \ldots , 10 \right \}$
the possible outcome is the number of the ball extracted, even if I don't know what is the order of that balls.

If I understand, if I pass to the function $X$, one of the element in $\Omega$, one of that $10$ balls, that function would retrieves the number "printed" on that particular ball.

$\begin{array}{lcl}P(X=0) & = & P(\left \{ \omega \in \Omega : X(\omega) = 0\right \}) \\ & = & P(\left \{ 1,2,3,4 \right \} ) \\ & = & \frac{4}{10} \end{array}$

$\begin{array}{lcl}P(X=1) & = & P(\left \{ \omega \in \Omega : X(\omega) = 1\right \}) \\ & = & P(\left \{ 5,6 \right \} ) \\ & = & \frac{2}{10} \end{array}$

$\begin{array}{lcl}P(X=2) & = & P(\left \{ \omega \in \Omega : X(\omega) = 2\right \}) \\ & = & P(\left \{ 7,8,9,10 \right \} ) \\ & = & \frac{4}{10} \end{array}$

hence,

$p_X(x) = \left \{ \begin{array}{rcl} \frac{4}{10} & \quad & 1 \le x \le 4 \\ \frac{2}{10} & \quad & 5 \le x \le 6 \\ \frac{4}{10} & \quad & 7 \le x \le 10 \\ 0 & \quad & x<1, x>10 \end{array} \right .$

I am really unsure about what I have done. Please, can you give me any suggestions on how to set up correctly this problem and correct any error? Many thanks!

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  • $\begingroup$ Your answer is fine except for the values that $X$ can take on. In your expression at the bottom you seem to be saying it can be any integer between 1 and 10, but it has to be a 0, 1 or 2. $\endgroup$ – dsaxton Feb 22 '17 at 14:32
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I think you're a bit off, your density function, for example, cannot be correct as

$$\sum_\omega p_X(\omega) = 4\cdot {4\over 10}+2\cdot {2\over 10}+4\cdot{4\over 10}$$

You have correctly identified that $\Omega = [10]$ with the uniform measure is the right call, however, the way the question is phrased you want $X(i)=`` \text{the number printed on ball }i\text{."}$ But you want $p_X(\omega)={1\over 10}$ for all $\omega$ because this is the density function, and all balls are equally likely, i.e. this is not about the probability of the value of $\omega$ it is the probability of picking a given ball. That is $P(X=X(\omega_0))\ne p_X(\omega_0)$ since $X^{-1}(X(\omega_0))$ is a full set which is probably bigger than just $\{\omega_0\}$.

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  • $\begingroup$ so the $x$ is one of the 10 balls. $$\begin{array}{rcl} x & 1 & \ldots & 10 \\ P(X=x) & \frac{1}{10} & \ldots & \frac{1}{10} \end{array}$$ but, what I have to do if I want to calculate the probability to draw a ball labelled with 2? $\endgroup$ – JB-Franco Feb 22 '17 at 18:45
  • $\begingroup$ @JB-Franco no, $p_X(x) = {1\over 10}$ here $x\in\Omega$ not $\{0,1,2\}$ remember $X:\Omega\to\Bbb R$. I think the notation is what is confusing you, in $p_X(x)$ we have $x\in\Omega$ (i.e. the domain) in $P(X=x)$ we have $x\in\Bbb R$, the target space, not the domain. $\endgroup$ – Adam Hughes Feb 22 '17 at 18:53
  • $\begingroup$ @JB-Franco I have edited my own notation to reflect more strongly where the argument lives, here $\omega$ is used to represent something in $\Omega$ to avoid confusion with values of $X$ which are in $\Bbb R$. $\endgroup$ – Adam Hughes Feb 22 '17 at 18:56
  • $\begingroup$ so the $P(X = x)$ I wrote in the first post is right, but it is wrong write the sets as $P(\left \{ 1,2,3,4 \right \})$ because there isn't any order. Only the part of $p_X(x)$ is wrong. $\endgroup$ – JB-Franco Feb 22 '17 at 19:13
  • $\begingroup$ @JB-Franco yeah, basically. The $p_X(x)$ is the density of an element of the domain whereas $P(X=x)$ is the probability you take a particular value in the range. What you really meant was $P(1\le X\le 4) = {4\over 10}$ which is a statement about the probability function, not about the probability density function. $\endgroup$ – Adam Hughes Feb 22 '17 at 19:21

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