8.6 Probability trees

Earlier modules gave us conditional probability for one known event and total probability for combining several routes. A probability tree is a way to organise those routes when an experiment unfolds in stages. Its branches show what can happen next given the path so far. This is especially useful when the second-stage chances change after the first result.

This is foundational to intermediate probability. OpenStax's probability-tree lesson explains branch multiplication, while its outcome-tree lesson helps with listing every path. Module 8.2 introduced a small replacement tree; here we use the tree as a complete calculation method.

A basic idea: every route has its own chance

A bag holds two red balls and one blue ball. Draw twice without replacement. The first draw is red with chance $2/3$ or blue with chance $1/3$. After red, the bag has one red and one blue, so the second draw has chances $1/2$ and $1/2$. After blue, both remaining balls are red, so the second draw is red with chance $1$.

For the path red then blue, multiply $P(RB)=(2/3)(1/2)=1/3$. For blue then red, $P(BR)=(1/3)(1)=1/3$. These paths cannot both occur, so the probability of one ball of each colour is $1/3+1/3=2/3$.

A quick tour of tree rules

Start at one root. At each node, draw all possible next outcomes and label their probabilities conditional on reaching that node; the labels leaving one node must add to $1$. Multiply labels along a complete path. Add the probabilities of separate complete paths that satisfy the event. All leaf probabilities together must total $1$ if the tree is complete.

For two stages, $P(A\text{ then }B)=P(A)P(B\mid A)$. The branch after $A$ carries $P(B\mid A)$, not necessarily the original $P(B)$. If the stage results are independent, that conditional chance stays the same. A tree can also be read backwards: to find the chance of a first-stage cause given a final result, divide the matching path total by the total probability of that result, as in Bayes' theorem.

Application 1: two draws without replacement

From a bag with two red balls and one blue ball, what is the chance that two draws without replacement have different colours?

1. The red-then-blue path has probability $(2/3)(1/2)=1/3$.
2. The blue-then-red path has probability $(1/3)(1)=1/3$.
3. Add the disjoint paths: $P(\text{different colours})=2/3$. The remaining path, red then red, has probability $1/3$, so the leaves total $1$.

Application 2: a coin chosen by a die

Roll a fair die. If it is even, use a coin with head probability $3/4$; if odd, use a coin with head probability $1/4$. Find the probability of a head and the chance the die was even given a head.

1. Even and odd each have probability $1/2$. The even-then-head path is $(1/2)(3/4)=3/8$; odd-then-head is $(1/2)(1/4)=1/8$.
2. Add them: $P(H)=3/8+1/8=1/2$.
3. Among head outcomes, the even path supplies $3/8$ of the total $1/2$, so $P(\text{even}\mid H)=(3/8)/(1/2)=3/4$.