Probability distribution--should these equal one?

**gobbajeezalus** · October 4th 2010, 08:34 AM

Hi there,

I'm supposed to prepare a probability distribution for the following question:

Two balls are drawn from a bag in which there are 4 white balls and 2 black balls. The number of black balls is being counted.

Okay.

So x = number of black balls.

Doing combinations:

Should those equal 1, or is it okay that they don't? I got .088 + .263 + .329, which doesnt = 1. Are my calculations correct?

Thanks!!

**Soroban** · October 4th 2010, 10:43 AM

Hello, gobbajeezalus!

I'm supposed to prepare a probability distribution for the following question:

Two balls are drawn from a bag which has 4 white balls and 2 black balls.
The number of black balls is being counted.

0 Black: . $\displaystyle {6\choose0}\!\left(\frac{1}{3}\right)^0\!\left(\fr ac{2}{3}\right)^6$

1 Black: . $\displaystyle {6\choose1}\left(\frac{1}{3}\right)^1\left(\frac{2 }{3}\right)^5$

2 Black: . $\displaystyle {6\choose2}\left(\frac{1}{3}\right)^2\left(\frac{2 }{3}\right)^4$

Are my calculations correct? . Sorry, no . . .

Your use of the "6" indicates that you are drawing all six balls from the bag.

Your use of $\tfrac{1}{3}$ and $\tfrac{2}{3}$ repeatedly indicates that
. . you are drawing the six balls with replacement.

I will assume the two balls are drawn without replacement.

$\displaystyle \text{Then there are: }\;{6\choose2} \:=\:15\text{ possible outcomes.}$

$\displaystyle P(\text{0B, 2W}) \;=\;\frac{{2\choose0}{4\choose2}}{15} \;=\;\frac{6}{15}$

$\displaystyle P(\text{1B, 1W}) \;=\;\frac{{2\choose2}{4\choose1}}{15} \;=\;\frac{8}{15}$

$\displaystyle P(\text{2B, 0W}) \;=\;\frac{{2\choose2}{4\choose0}}{15} \;=\;\frac{1}{15}$

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