Probability/Combinatorics: # ways of picking 5 from 3 groups of 6

[Master1022]

Homework Statement

We have a bag of 18 marbles: 6 red, 6 blue, and 6 yellow. Now I randomly select 5 marbles from the bag without replacement. What is the probability that I have picked out EXACTLY 2 colors?

Relevant Equations

Combinatorics

Hi,

I was attempting the following question and was getting slightly stuck.

Question: We have a bag of 18 marbles: 6 red, 6 blue, and 6 yellow. Now I randomly select 5 marbles from the bag without replacement. What is the probability that I have picked out EXACTLY 2 colors?

Attempt: I tried to think about this problem in two different ways:

1) Pure counting argument such that $p = \frac{\text{Number of ways which we pick 2 colors}}{\text{Total number of ways of picking 5 from 18}}$

So my logic was as follows:
- there are $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ ways of picking 2 out of the three colors
- Then for each of those pairs of colors (let us call them A and B), we can do: (1 from A, 4 from B), (2 from A, 3 from B), (3 from A, 2 from B), (4 from A, 1 from B). This can be written more formally as:
\begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix}

and thus this becomes:
\begin{pmatrix} 3 \\ 2 \end{pmatrix} \left( \begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} \right)

I am slightly confused on how to get the total number of ways of picking 5 from 3 groups of 6. I mean I can see the obvious $\begin{pmatrix} 18 \\ 5 \end{pmatrix}$, but doesn't that double count some groupings?? This was asked as an interview question, so I don't think I would have time to write down very elaborate alternative methods.

This would lead to:
p = \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \left( \begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} \right)}{\begin{pmatrix} 18 \\ 5 \end{pmatrix}} = \frac{65}{238}

2) The other 'method' was to frame the problem like: "how many solutions are there to the equation $x_1 + x_2 = 5$ where $x_1 \geq 1$ and $x_2 \geq 1$.
- so there are still $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ ways of picking 2 out of the three colors
- then there would be $\begin{pmatrix} (5 - 2) + (2 - 1) \\ (2 - 1) \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ ways choosing the groups...

- For the total number of ways to pick 5 from 18, we could use a similar framing of $x_1 + x_2 + x_3 = 5$, but instead just have $x_1 , x_2, x_3 \geq 0$. This leads to $\begin{pmatrix} 5 + (3 - 1) \\ (3 - 1) \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$

This would lead to:
p = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 7 \\ 2 \end{pmatrix}} = \frac{4}{7}

Can anyone help to reconcile which one of these methods is more appropriate for this problem?

Many thanks.

 

[PeroK]

I would simply do:$$P = 3P(R+B) = 3(P(RB) - 2P(R))$$I'll let you work out the notation!

 

[Master1022]

I would simply do:$$P = 3P(R+B) = 3(P(RB) - 2P(R))$$I'll let you work out the notation!

Thanks @PeroK ! Unfortunately, I have been looking at this for a while and am a bit confused. Here is what I have made of it. We are using the probability rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

You have P(R + B) as the $P(R \cup B)$, but then I don't understand why the final term in the equation is $P(R \cap B) - 2 \cdot P(R)$ instead of $2 P(R) - P(R \cap B)$? (is that a correct understanding?

 

[PeroK]

Thanks @PeroK ! Unfortunately, I have been looking at this for a while and am a bit confused. Here is what I have made of it. We are using the probability rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

You have P(R + B) as the $P(R \cup B)$, but then I don't understand why the final term in the equation is $P(R \cap B) - 2 \cdot P(R)$ instead of $2 P(R) - P(R \cap B)$? (is that a correct understanding?

Well, $R + B$ must indicate one of the events you are looking for; namely, only Red and Blue balls and at least one of each. I'm not sure I understand the idea of translating to set notation? Why not leave things as they are?

Do you understand the factor of $3$?

I didn't want to make it too easy, so I wrote $-2P(R)$ instead of the more suggestive $-P(R) - P(B)$.

 

[PeroK]

PS I just checked the answer: $65/238$ is correct.

 

[Master1022]

Well, $R + B$ must indicate one of the events you are looking for; namely, only Red and Blue balls and at least one of each.

Oh I was thinking in terms of Venn diagrams, when R + B (with + being a union), but I guess that isn't what you meant? Then what does (RB) represent - I would have thought that meant both red and blue colored balls.

I'm not sure I understand the idea of translating to set notation? Why not leave things as they are?

Hmm, that is what came to mind when I saw the form of that equation...

Do you understand the factor of $3$?

I think that is from the 3C2 = 3 (i.e. the number of ways to choose 2 colors from 3)

I didn't want to make it too easy, so I wrote $-2P(R)$ instead of the more suggestive $-P(R) - P(B)$.

I understood the symmetry part of it ($P(R) = P(B)$) so we can combine them. I think my question was why the equation was $P(RB) - 2P(R)$ instead of the other way around (i.e. $2P(R) - P(RB)$)

 

[PeroK]

I understood the symmetry part of it ($P(R) = P(B)$) so we can combine them. I think my question was why the equation was $P(RB) - 2P(R)$ instead of the other way around (i.e. $2P(R) - P(RB)$)

I guess I should come clean:

$R + B$ was only red and blue with at least one of each.

$RB$ was only red and blue (i.e. no yellows) and includes the cases of all red and all blue. I guess I could have used $no \ Y$ instead.

$R$ is all red.

 

[Master1022]

Actually the 18C5 makes sense if I think about the Vandermonde Identity.