Probability Question - Ball Urn Problem

[GreenPrint]

Using the binomial theorem you can solve ball urn problems. Like say for example a urn has 4 green balls and 3 white balls. You draw two balls. What's the probability you draw 2 white balls?

I just made this problem up off the top of my head. But anyways using the binomial theorem

(7C2)(3/7)^2*(4/7)^5

However this is wrong correct? Because it's impossible to draw two balls simultaneously... I just realized that. So how do you solve this problem then?

 

[Ray Vickson]

Using the binomial theorem you can solve ball urn problems. Like say for example a urn has 4 green balls and 3 white balls. You draw two balls. What's the probability you draw 2 white balls?

I just made this problem up off the top of my head. But anyways using the binomial theorem

(7C2)(3/7)^2*(4/7)^5

However this is wrong correct? Because it's impossible to draw two balls simultaneously... I just realized that. So how do you solve this problem then?

The issue is one of drawing *with* replacement (binomial applies) vs. drawing *without* replacement (binomial fails). For drawing without replacement you need to use the hypergeometric distribution.

In your case, the probability that the first drawn ball is W is 3/7. Given that the first one is W that leaves 6 balls in the urn, of which 2 are white, so the probability the next one drawn is W is 2/6. The probability of drawing 2 whites is P{WW} = (3/7)*(2/6) = 1/7.

For more on the hypergeometric distribution see
http://en.wikipedia.org/wiki/Hypergeometric_distribution
or
http://mathworld.wolfram.com/HypergeometricDistribution.html
or
http://www.math.uah.edu/stat/urn/Hypergeometric.html .

 

[GreenPrint]

Well what if you don't know if the first ball is drawn is W or or G?

1/7 if you draw W on the first draw
but if you draw R on the first than the probability you draw two W is zero

so how would you answer this question then since there are two answers I guess?

 

[Ray Vickson]

Well what if you don't know if the first ball is drawn is W or or G?

1/7 if you draw W on the first draw
but if you draw R on the first than the probability you draw two W is zero

so how would you answer this question then since there are two answers I guess?

The answer is the same, whether or not you *know* the color of the balls.

 

[GreenPrint]

Why is that? If you draw a Red ball on the first than there's no way to draw two white balls.

 

[Ray Vickson]

Why is that? If you draw a Red ball on the first than there's no way to draw two white balls.

You asked for the probability of two white balls.

 

[GreenPrint]

So given the question how do you know if there's replacement or not?

 

[Ray Vickson]

So given the question how do you know if there's replacement or not?

You don't. Somebody needs to tell you, or you need more context. For example, in sampling industrial output by destructive testing, there cannot be any possibility of replacement.