Gumballs

eulB

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Jan 20, 2005
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I need help. (Duh dosent every one) The problem goes theres a gumball machine with equal amounts of red blue and white gumballs. What is the probablity of geting a red white and blue gum ball. THE TEACH SAYS to find out the problem you need to write down every possible combanation of gumball you could get. I think i have the answer but isnt there an equation that could do the job faster and with more accuratecy (thats so spelled wrong oh wellz)
 
The problem is very poorly defined. It is a very small wonder that you are struggling with it.

Are we making three (3) draws? It doesn't say so. It should. I'll assume that is the case.
In that order? Will Blue,Red,White count? It doesn't say so. It should. I'll assume that is the case.
Do we put back the gumballs we pick or do we just go on? It doesn't say so. It should. I'll assume that we don't put things back between draws.

"equal amounts" is also very much insufficient.

If "equal amounts" means zero (0) each, then Pr(Red,White,Blue) = 0.

If "equal amounts" means one (1) each, then Pr(Red,White,Blue) = 1.

If "equal amounts" means some relative small number, then Pr(White|Red) can be significantly different from Pr(White). Let's try 2. Pr(Red,White,Blue) = 0.40. I won't bother to tell you how I found that.

If "equal amounts" means some really large number, then we have a TRINOMIAL DISTRIBUTION for which all probabilities can be calculated easily. Expand (R + W + B)<sup>3</sup> to get:

R<sup>3</sup>+3*R<sup>2</sup>*W+3*R<sup>2</sup>*B+3*R*W<sup>2</sup>+6*R*W*B+3*R*B<sup>2</sup>+W<sup>3</sup>+3*W<sup>2</sup>*B+3*W*B<sup>2</sup>+B<sup>3</sup>

R = Pr(Red on any draw) = 1/3
W = Pr(White on any draw) = 1/3
B = Pr(Blue on any draw) = 1/3

Picking out the term with all three: 6*R*W*B ==> 6*(1/3)<sup>3</sup> = 6/27 = 2/9 = 0.222222...

How close were you? How close am I?

With a problem statement like this, I'm afraid I would have to prevail upon the teacher to accept any reasonable set of assumption and an answer consistent with those assumptions.
 
Hope this is correct...not disputing TK's; just assuming different:
1: one of each color on 1st 3 draws
2: any order, no replacement

gumballs = g = b + r + w where b = r = w

pick 1st [assume b]
pick 2nd: r or w probability = (r + w) / (g - 1) [assume r]
pick 3rd: w probability = w / (g - 2)

So probability = [(r + w) / (g - 1)] * [w / (g - 2)]
 
gumballs

Hello, eulB!

Your teacher's advice is correct for 'small' problems with specific numbers given,
. . but not for this one.
TK is absolutely correct. . The problem is egregiously sloppy.
. . The original wording would be helpful.

There are 3n gumballs: n each of the colors red, white, blue.

Assume that <u>three</u> gumballs are drawn <u>without</u> replacement, and n > 1.

Specify an <u>order</u> to the colors, say: R-W-B.
. . Pr(1st is R) = n/3n = 1/3
. . Pr(2nd is W) = n/(3n - 1)
. . Pr(3rd is B) = n/(3n - 2)
. . . . . . . . . . . . . . . . . . . . . . . . n<sup>2</sup>
Hence: . Pr(R-W-B) .= .---------------------
. . . . . . . . . . . . . . . . . . 3(3n - 1)(3n - 2)

Since the order is not important, there are 3! = 6 possible orders.
. . Multiply our result by 6.
. . . . . . . . . . . . . . . . . . . . . . . 2n<sup>2</sup>
Then: . Pr(R, W, B) . = . --------------------
. . . . . . . . . . . . . . . . . . . (3n - 1)(3n - 2)

Note: As n increases, the probability approaches 2/9.
 
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