- #1

Master1022

- 611

- 117

- 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.

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:

[tex] \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} [/tex]

and thus this becomes:

[tex] \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) [/tex]

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:

[tex] 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} [/tex]

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:

[tex] p = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 7 \\ 2 \end{pmatrix}} = \frac{4}{7} [/tex]

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

Many thanks.

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:

[tex] \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} [/tex]

and thus this becomes:

[tex] \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) [/tex]

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:

[tex] 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} [/tex]

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:

[tex] p = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 7 \\ 2 \end{pmatrix}} = \frac{4}{7} [/tex]

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

Many thanks.