Three green balls without replacement

Problem

An urn contains $3$ red balls and $5$ green balls. Three balls are drawn without replacement. Given that the last one is green, what is the probability that all three are green? Hint: build the full three-level tree and apply the conditional probability formula.

Solution

Let $V$ be the event «the third drawn is green» and $VVV$ the event «all three green». We seek $P(VVV\mid V)=\dfrac{P(VVV)}{P(V)}$.

Denominator $P(V)$: summing over the tree all the paths that end with green, $P(V) = \frac{5}{8}\frac{4}{7}\frac{3}{6} + \frac{5}{8}\frac{3}{7}\frac{4}{6} + \frac{3}{8}\frac{5}{7}\frac{4}{6} + \frac{3}{8}\frac{2}{7}\frac{5}{6} = \frac{5}{8}.$ (As expected: by symmetry the third ball is green with the same probability $5/8$ as the first.)

Numerator $P(VVV)$: a single path, $P(VVV) = \frac{5}{8}\frac{4}{7}\frac{3}{6} = \frac{5}{28}.$ Therefore $P(VVV\mid V) = \frac{5/28}{5/8} = \frac{8}{28} = \boxed{\dfrac{2}{7}}$

Links

Topics: Probabilita
Concepts: Diagramma albero · Probabilita condizionata
Methods: Prob condizionata
Skills: Calcolo probabilita
Exercise type: Problema probabilita