Hypergeometric — urn with 3 red and 5 green

This is the guiding example that lets us compare two equivalent approaches: the probability tree (introduced in Year 4) and the compact hypergeometric formula.

Example — Urn with 3 red and 5 green

An urn with $8$ balls: $3$ red and $5$ green. You draw $3$ balls without replacement. What is the probability of getting exactly $2$ red?

With $N=8$, $K=3$, $n=3$, $k=2$: $P(X=2) = \frac{\binom{3}{2}\binom{5}{1}}{\binom{8}{3}} = \frac{3\cdot 5}{56} = \frac{15}{56}\approx 0{,}268.$

The same value is obtained with the tree: there are $\binom{3}{2} = 3$ possible orders of appearance ($RRV, RVR, VRR$), each with probability $\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{5}{6} = \frac{30}{336} = \frac{5}{56};$ their sum is $15/56$. The hypergeometric formula compacts this reasoning into a single step. The mean is $\mu = 3\cdot\frac{3}{8} = \frac{9}{8} \approx 1{,}125 \text{ expected red}.$

Links

Topics: Probability distributions
Concepts: Hypergeometric distribution · Expected value
Methods: Hypergeometric distribution
Skills: Combinatorics · Probability calculation