# Urn problem: 2 balls randomly without replacement.

#### math951

we have an urn with 3 green and 4 yellow balls. We choose 2 balls randomly without replacement. Let A be the event that we have two different color balls in our sample.

Describe a sample space: sample space: (G,Y), (Y,G), (G,G), (Y,Y)
Describe event A = (G,Y) and (Y,G)

P(A) = (2 * 4C1 * 3C1)/ 7C2

Did I do this right?

#### romsek

MHF Helper
Your probability has an extra factor of 2. You don't need to account for both cases when using combinations.

$$\displaystyle P(A) = \dfrac{\dbinom{3}{1}\dbinom{4}{1}}{\dbinom{7}{2}}$$

#### math951

Just to be clear, if I do an experiment, and I sample without replacement, order matters, then that would give me a different answer versus doing the experiment where I sample without replacement and order does not matter.

#### Plato

MHF Helper
we have an urn with 3 green and 4 yellow balls. We choose 2 balls randomly without replacement. Let A be the event that we have two different color balls in our sample.
Describe a sample space: sample space: (G,Y), (Y,G), (G,G), (Y,Y)
Describe event A = (G,Y) and (Y,G)
P(A) = (2 * 4C1 * 3C1)/ 7C2
Did I do this right?
Just to be clear, if I do an experiment, and I sample without replacement, order matters, then that would give me a different answer versus doing the experiment where I sample without replacement and order does not matter.
In this space there are four elementary events: $(G,Y), (Y,G), (G,G), (Y,Y)$
$$\displaystyle \mathcal{P}((G,Y))=\dfrac{3}{7}\cdot\dfrac{4}{6}=\dfrac{2}{7}$$ BUT ALSO $\mathcal{P}((Y,G))=\dfrac{4}{7}\cdot\dfrac{3}{6}=\dfrac{2}{7}$
Moreover $\dfrac{\dbinom{3}{1}\dbinom{4}{1}}{\dbinom{7}{2}}=\dfrac{4}{7}$ Did order matter?

Last edited:
chip and math951

#### math951

@Plato I guess my question is when do we know which sampling w/out replacement to do?

I have typically seen if we take one out one by one, to use sampling without replace, order matters.

If not, then do sample without replacement, order irrelevant.

#### Plato

MHF Helper
I guess my question is when do we know which sampling w/out replacement to do?
I have typically seen if we take one out one by one, to use sampling without replace, order matters.
If not, then do sample without replacement, order irrelevant.
If the question is about content without replacement the order makes no difference. That is the case in this question.
If from a group of six men and eight women we randomly select a committee of five.
What is the probability of getting three men and two women: $\dfrac{\binom{8}{2}\binom{3}{6}}{\binom{5}{14}}$ No order.
From the whole of fourteen at random we pick s Pres, VP, Sec & Treasure.
There are ${\mathscr{P}^{14}_4}=\dfrac{14!}{(14-4)!}=24024$ permutations ways to do that.
How many ways for the there to be $(W,M,W,M)~?$
How many ways for the there to be (Ellen,Mark,Jane,Max)?
The probability of one is $\dfrac{1}{24024}$ and of the other is $\dfrac{\mathscr{P}^{8}_2\mathscr{P}^{6}_2}{\mathscr{P}^{14}_4}$. Which is which & WHY?

#### math951

(W,M,W,M) would be the second one because we are focused on arrangements of 2 women 2 men..
(ellen, mark, jane, max)... because 14P4 is 24024.. the probability of an event would be P(w)= (24024)^-1...
Hence (ellen,mark,jane,max) is just the probability of an event.