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In a urn with coins there are: $3$ of $1$ dollars, $2$ of $0.5$ dollars and $1$ of $0.1$ dollars.

All coins are identical and can not be distinguished by touch.

Two coins are drawn without looking: a first and a second (without replacing the first).

Let $X$ be the random variable that collects the gain of the game:

  1. What kind of variable is $X$?
  2. Find the density or quantity function
  3. Find $\mathcal{P}(X = 1)$

I know that a discrete random variable must have a finite or infinite numerable range.

So I can count the amount of values and possible gains by adding two coins:

Let $Y$ "The value of the coin",

$\mathcal{P}(Y=0.1)=1/6$

$\mathcal{P}(Y=0.5)=1/3$

$\mathcal{P}(Y=1)=1/2$

-Assuming X is discrete-

  • $(0.1+0.5)=(0.5+0.1)=0.6\Rightarrow \mathcal{P}(X=0.6)=(1/6)(1/3)+(1/3)(1/6)=1/9$
  • $(0.5+0.5)=1\Rightarrow \mathcal{P}(X=1)=(1/3)(1/3)=1/9$
  • $(1+0.1)=(0.1+1)=1.1\Rightarrow \mathcal{P}(X=1.1)=(1/2)(1/6)+(1/6)(1/2)=1/6$
  • $(1+0.5)=(0.5+1)=1.5\Rightarrow \mathcal{P}(X=1.5)=(1/2)(1/3)+(1/3)(1/2)=1/3$
  • $(1+1)=2\Rightarrow \mathcal{P}(X=2)=(1/2)(1/2)=1/4$

But, what I did is right?

Thank you very much.

1 Answers 1

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It is clear that $X$ is discrete since it can only take a finite amount of values.

Your solution for the pmf is not correct, however, because you are doing draws with replacement.

We can let $Y_1$ be the first draw and $Y_2$ the second. Then you have $$ \mathbb P(X = 0.6) = \mathbb P((Y_1 = 0.5, Y_2 = 0.1) \cup (Y_1 = 0.1, Y_2 = 0.5)) \\ = \mathbb P(Y_1 = 0.5) \mathbb P(Y_2 =0.1 \mid Y_1 = 0.5) + \mathbb P(Y_1 = 0.1) \mathbb P(Y_2 =0.5 \mid Y_1 = 0.1) \\ = \frac{2}{6} \frac{1}{5} + \frac{1}{6} \frac{2}{5} $$ and similar for the other combinations.