Covariance of red and blue counts drawn without replacement

Question

A bag holds 10 balls: 4 red, 3 blue, and 3 green. You draw 5 balls without replacement. Let NR be the number of red drawn and NB the number of blue. What is the covariance of NR and NB?

Accepted Answer

Two forces both push this covariance negative. First, the counts compete for the same 5 draws (a slot filled by red cannot be blue). Second, drawing without replacement means pulling one color early leaves fewer of everything for later. For the multi-color draw-without-replacement setup, the covariance of two different color counts is \mathrm{Cov}(NR, NB) = -\,n \cdot \frac{KR}{N} \cdot \frac{KB}{N} \cdot \frac{N - n}{N - 1}, where N is the total balls, KR and KB are how many of each color, and n is the number drawn. The first three factors mirror the with-replacement formula -n\,p\,q; the fi…

Covariance of red and blue counts drawn without replacement

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