Probability, where am I going wrong?

May 2010
7
0
An urn contains 10 white and 5 red marbles. Three marbles are selected at random without replacement from the urn. Set up the probability distribution for the randomn variable X defined as the number of red marbles selected. Determine the expected value and the variance of X.

So what I did

P(WWW) = 0.2637
P(WWR) = 0.1648
P(WRW) = 0.1648
P(WRR) = 0.0733
P(RWW) = 0.1648
P(RWR) = 0.0733
P(RRW) = 0.0733
P(RRR) = 0.0220

The mean is 1 and the variance is 0.5056.

My problem is, I keep on getting 0.5714, where am I going wrong?
 

Plato

MHF Helper
Aug 2006
22,462
8,634
I think that you are going about in the hard way.
\(\displaystyle P(R=x)=\frac{\binom{10-x}{3-x}\binom{5}{x}}{\binom{15}{3}}\), That is the probability of having \(\displaystyle x=~0,~1,~2,\text{ or }3\) red balls in the sample.

BTW: I get the same variance as you.
 
Last edited:

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
This is a hypergeometric distribution.
You can check the mean (1) and variance at...http://en.wikipedia.org/wiki/Hypergeometric_distribution

The mean is the same as with the binomial (np), even though this is sampling wor.
n=3 and p=5/15=1/3 giving you E(X)=1.
The variances are slightly different, but as the lot sizes increase they approach each other.