Doubt about random variables and sample spaces

Question

I have this little problem, reviewing my notes from the university of the first days of probability, I realized that there was a blank problem in my notes, says the following:

" In a bag there are 5 balls, 3 red and 2 white, for some reason we are interested in taking 3 out of the bag at random"

as in most of these problems we had to find the sample space, define some random variable $W$ which represents the number of white balls, build the probability distribution etc etc

The first thing I thought is that, it seemed that it would be solved using a hypergeometric distribution, it made sense, what causes me conflict is a note on the side that the sample space was written down

$\Omega= \lbrace RRR,RRW,RWR,RWW,WRR,WRW,WWR,WWW \rbrace$ which I imagine $R$ is red and $W$ white respectively

If we are interested in $ X =$ the number of white balls in the selection of 3 balls and since $ WWW $ cannot happen because there are only two white balls in the bag, the probability distribution should be

$P(X=0)=\frac{1}{7}$, $P(X=1)=\frac{3}{7}$, $P(X=2)=\frac{3}{7}$

this is correct?? when I attacked it by the hypergeometric it gave me these results $\frac{1}{10},\frac{3}{10},\frac{6}{10}$

Answer 1

You didn't provide a few key pieces of information in your question (such as if ordering matters). However, I'll pick up on the "early days of probability" quip and try to articulate a version of the problem and the solution. Let us label the red and white balls in the bag as $\{R_1, R_2, R_3, W_1, W_2\}$. The sample space you listed is not correct since $\omega = \{WWW\}$ cannot be an element since there are only two white balls.

If the ordering does not matter, then your sample space contains 10 combinations: $$\Omega = \{R_1 R_2 R_3, R_1 R_2 W_1, R_1 R_2 W_2, R_1 R_3 W_1, R_1 R_3 W_2, R_2 R_3 W_1, R_2 R_3 W_2, R_1 W_1 W_2, R_2 W_1 W_2, R_3 W_1 W_2 \}$$

So in this scenario, if $X$ is the number of white balls selected with each draw of three balls, $P(X = 0) = 1/10$, $P(X = 1) = 6/10$, and $P(X = 2) = 3/10$.

I hope that helps.

Answer 2

First of all, it's way to early to talk about any distributions yet, hypergeometrix or not. Finding the sample space has nothing to do with that. The sample space describes the outcomes of an experiment, regardless of what we're going to count in this experiment. Say, in this experiment (and we'll have to talk more about what "this experiment" even means ) we can define a random variable $X=\text{[number of red balls]}$ or a random variable $Y=\text{[number of white balls]}$ — and they are two different things with different distributions, even though the underlying sample space is the same.

Now, what is "this experiment"? Two common scenarios of drawing from a set (bin) of items are:

• drawing without replacement, where after you draw each item you keep it to yourself;

• and drawing with replacement, where after you draw each item you return it to the bin.

Judging by your notes, this was the drawing with replacement scenario, for which the sample space is precisely as listed in your notes.

Of course, in the drawing without replacement scenario your logic would be perfectly correct, and the outcome $WWW$ is impossible in that case.

EDIT: One more thing I forgot to mentioned before I submitted my answer at first. The way you calculate thes probabilities is also different for drawing with or without replacement. Hypergeometric distribution applies only to one of them.

Answer 3

The outcomes in this sample space are not equally likely. Even though we might not care which of the white balls and which of the black balls we choose, in principle they are distinguishable, and this affects the probabilities. If we label the white balls $W_1$ and $W_2$ and the red balls $R_1, R_2, R_3$, then for example your $RWW$ corresponds to $R_1 W_1 W_2$ or $R_2 W_1 W_2$ or ... or $R_3 W_2 W_1$, $6$ possibilities, while $RRW$ would have $12$ possibilities.