Urn of 12 balls, mean and variance

Question

From an urn with $N=12$ balls ($K=6$ white, $6$ black) $n=6$ balls are drawn without replacement. Compute the mean and variance of $X$ = number of white balls and compare them with the “with replacement” case ($B(6;\,1/2)$). Comment on the correction factor.

Solution

Hypergeometric $\text{Hyp}(12,6,6)$: $\mu = n\cdot\frac{K}{N} = 6\cdot\frac{6}{12} = \boxed{3},$ $\sigma^2 = n\cdot\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1} = 6\cdot\frac12\cdot\frac12\cdot\frac{6}{11} = 1{,}5\cdot\frac{6}{11}\approx \boxed{0{,}818}.$ Binomial $B(6;\,1/2)$ (with replacement): $\mu = np = 3$ (same mean), $\sigma^2 = np(1-p) = 6\cdot\tfrac12\cdot\tfrac12 = 1{,}5$.

The two means coincide; the hypergeometric variance is smaller because of the finite-population correction factor $\dfrac{N-n}{N-1} = \dfrac{6}{11}\approx 0{,}545$. Without replacement the results are less spread out: drawing half the urn strongly constrains the outcome (if many white balls come out, few remain), reducing the variability.

Links

Topics: Probability distributions
Concepts: Hypergeometric distribution · Expected value · Variance
Methods: Hypergeometric distribution
Skills: Limiting-case analysis · Probability calculation
Exercise type: Probability problem