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

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Homework Help Overview

The problem involves calculating probabilities related to car orders in four colors: blue, white, green, and black. The specific questions focus on the likelihood of ordering one blue, one white, and one green car, as well as the probability of having exactly two orders of the same color.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the total number of combinations for car orders and question whether different arrangements of the same colors should be considered unique or identical. There is also consideration of the assumptions regarding the probabilities of each color being ordered.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying the assumptions about color probabilities. Some guidance has been offered regarding the calculation of combinations and the significance of order in the context of the problem.

Contextual Notes

There is uncertainty regarding the assumptions about the likelihood of each color being ordered, as this is not explicitly stated in the problem. Participants are also grappling with the implications of treating different arrangements of the same color combination as either unique or identical.

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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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