Urn of 10 balls, complete distribution

Question

An urn has $10$ balls, of which $4$ are red. Three balls are drawn without replacement: compute $P(X=0), P(X=1), P(X=2), P(X=3)$ and verify that the sum is $1$.

Solution

Hypergeometric model with $N=10$, $K=4$, $n=3$; the denominator is $\binom{10}{3}=120$. $P(X=0) = \frac{\binom{4}{0}\binom{6}{3}}{120} = \frac{20}{120} = \frac{1}{6}\approx 0{,}167,$ $P(X=1) = \frac{\binom{4}{1}\binom{6}{2}}{120} = \frac{4\cdot 15}{120} = \frac{60}{120} = \frac{1}{2},$ $P(X=2) = \frac{\binom{4}{2}\binom{6}{1}}{120} = \frac{6\cdot 6}{120} = \frac{36}{120} = 0{,}3,$ $P(X=3) = \frac{\binom{4}{3}\binom{6}{0}}{120} = \frac{4}{120} = \frac{1}{30}\approx 0{,}033.$ Check: $20+60+36+4 = 120$, hence $\sum_k P(X=k) = \dfrac{120}{120} = \boxed{1}$. ✓

Links

Topics: Probability distributions
Concepts: Hypergeometric distribution
Methods: Hypergeometric distribution
Skills: Combinatorial calculation · Probability calculation
Exercise type: Probability problem