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Angela_o
07-29-2011, 02:55 AM
Please can someone help me with the following question:

An urn contains 3 white, 6 red and 5 black balls. Six of these balls are randomly selected from the Urn. Let X and Y denote respectively the number of white and black balls selected. Compute the conditional probability mass function of X given that Y = 3. Also compute E[X|Y = 1].

I am trying to compute the probability of the various combinations of the balls that are drawn from the urn. Thus P(0, 1, 5) = 0,0099 by (6¦5) {(5¦1)(14¦6)} (the are combinations of the form n combination x). I am struggling though to calculate the rest of the probability combinations. If I wanted to calculate P(1, 0, 5) what would the combination be to calculate the probability like the combination (6¦5) {(5¦1)(14¦6)} was used to calculate P(0, 1, 5).

I hope someone can help me, been struggling with this for days now. Thanks!

BGM
08-08-2011, 11:16 AM
Let Z be the number of red balls selected. Then (X, Y, Z) jointly follow the multivariate hypergeometric distribution, i.e.

\Pr\{X = x, Y = y, Z = z\} = \frac {\displaystyle \binom {3} {x} \binom {5} {y} \binom {6} {z}} {\displaystyle \binom {14} {6}}

Note that you have the constraint X + Y + Z = 6 . Hence

\Pr\{X = x|Y = 3\} = \frac {\Pr\{X = x, Y = 3\}} {\Pr\{Y = 3\}} = \frac {\Pr\{X = x, Y = 3, Z = 3 - x\}} {\Pr\{Y = 3\}}
x = 0, 1, 2, 3

And the marginal distribution is just the hypergeometric distribution. So you can compute the conditional probability mass function easily, and the corresponding expectation as well.