# Probability Question

1. Jul 26, 2010

### Tangent...

1. The problem statement, all variables and given/known data
A firm sells a car in four colors; blue, white, green, and black. Three successive orders are placed for the automobile. What is the probability that one blue, one white, and one green are ordered? Exactly two of the orders are the same color?

2. Relevant equations

3. The attempt at a solution
I first tried finding the total number of ways three orders could be placed with the color combos. So I used the combination formula for when repeats are possible, "(n+r-1) choose r" and came up with 36 possible order scenarios. I then figured that there is only one possible way to get the combo blue, white, and green (since order does not matter?). I got 1/36 which is not the answer. Any help would be great, I'm not sure what I'm missing, but I'm definitely missing something.

2. Jul 27, 2010

### hgfalling

There may be 36 different combinations, but not all of them are equally probable, right?

WGU, WUG, GUW, GWU, UGW, UWG are all the same combination, but there's only one way to make BBB.

3. Jul 27, 2010

### Tangent...

I was under the impression that WGU/WUG/GUW/etc. was considered a single combination, should I be thinking of each one as unique?

4. Jul 27, 2010

### hgfalling

Well what are your assumptions about the likelihoods of the colors of the orders? I mean, it seems natural to me from the way these types of problems are usually posed that each order is a random color. That's not specified in your problem, however, so maybe I shouldn't assume that.

But if the answer to the first part of the question is 3/32, then my assumption is correct. (edited)

Maybe this will explain what I mean. When you roll two dice and add them up, there are 11 different totals they could be. But the probability that the total is 9 isn't 1/11.

So UWG and UGW are the same combination. But that combination isn't equally likely with the others.

5. Jul 27, 2010

### Tangent...

The answer is 3/32, so it looks like your assumptions are correct. Just to clarify, in this case the order does matter, so UWG is different than WGU? How do we arive at the total number of options, since 36 doesn't seem to fit?

6. Jul 27, 2010

### hgfalling

Well, there are four colors. So each order can be one of the four colors. So 43=64.

Six of those make the WGU combo, so it's 6/64 = 3/32.

7. Jul 27, 2010

### Tangent...

I'll believe that. Thank you so much for the help!