Probability of Blue, White, Green Car Orders & Exactly 2 Same Color

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The discussion revolves around calculating the probability of specific car color orders from a firm that offers blue, white, green, and black cars. The initial attempt to find the probability of ordering one blue, one white, and one green resulted in confusion regarding the total number of combinations. It was clarified that the order of colors matters, leading to a total of 64 possible combinations for three orders. The correct probability for the combination of one blue, one white, and one green is determined to be 3/32, confirming that the assumptions about random color selection were valid. The conversation highlights the importance of understanding how order affects probability calculations in combinatorial scenarios.
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Homework Statement


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?


Homework Equations





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.
 
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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.
 
I was under the impression that WGU/WUG/GUW/etc. was considered a single combination, should I be thinking of each one as unique?
 
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.
 
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?
 
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.
 
I'll believe that. Thank you so much for the help!
 
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