Tree Diagrams

One set of branches for each stage

Toss a fair coin twice. A tree diagram draws the first toss as two branches from a starting point, H for a head and T for a tail, each with its probability, $\frac{1}{2}$1/2, written on it.

From the end of each first branch, the second toss splits again into H and T. So the tree has 2 × 2 = 4 paths from the start to the right-hand side: HH, HT, TH and TT. Each path is one outcome of the two tosses.

The branches from any one point add to 1, because one of them must happen: $\frac{1}{2}+\frac{1}{2}=1$1/2 + 1/2 = 1.

The first toss on the left, the second toss after each of its branches, and the four outcomes down the right.

Multiply along a path

The probability of two heads is the probability of the path H then H. Half of all pairs of tosses start with a head, and half of those go on to a second head, so the chance is $\frac{1}{2}$1/2 of $\frac{1}{2}$1/2: $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$1/2 × 1/2 = 1/4.

Multiplying along a path works because the tosses are independent: the second toss has the same chances whatever the first one did. Every path here is $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$1/2 × 1/2 = 1/4, and the four paths add to 1, because the two tosses must end on exactly one of them.

The path H then H is colored: $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$1/2 × 1/2 = 1/4. Each of the four paths ends on $\frac{1}{4}$1/4, and $4\times \frac{1}{4}=1$4 × 1/4 = 1.

Branches that are not equal

A basketball player scores a free throw with probability 0.7, so misses with probability 0.3, and her two throws are independent. On the tree, each throw splits into "in", 0.7, and "out", 0.3, and the pair after the first throw is the same whichever way it went.

Multiply along each path: in then in is 0.7 × 0.7 = 0.49, in then out is 0.7 × 0.3 = 0.21, out then in is 0.3 × 0.7 = 0.21, and out then out is 0.3 × 0.3 = 0.09. Check: 0.49 + 0.21 + 0.21 + 0.09 = 1.

A grid of equally likely cells could not draw this, because a score and a miss are not equally likely. A tree writes each chance on its branch, so it works for any chances.

Two free throws, each scored with probability 0.7. The path in then in is colored, and the chance at the end of each path is the product of its two branches.

Adding paths

What is the probability that she scores exactly one of the two throws? Two paths do that: in then out, and out then in. A pair of throws follows only one path, so the two paths are mutually exclusive, and their chances add: 0.21 + 0.21 = 0.42.

The three answers "both", "exactly one" and "neither" cover every case: 0.49 + 0.42 + 0.09 = 1.

The usual mistakes

Adding along a path instead of multiplying. For two heads, $\frac{1}{2}+\frac{1}{2}=1$1/2 + 1/2 = 1 would make two heads certain. Two wins in a row with $\text{a}\frac{1}{3}$a 1/3 chance each is $\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$1/3 × 1/3 = 1/9, not $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$1/3 + 1/3 = 2/3.

Adding the denominators: $\frac{1}{\left(3+3\right)}=\frac{1}{6}$1/(3 + 3) = 1/6. Taking $\frac{1}{3}$1/3 of $\frac{1}{3}$1/3 cuts each third into 3 parts, so there are 3 × 3 = 9 equal parts.

Counting only one path when two give the answer, such as in then out for "exactly one". Out then in is a different path to the same answer.

Writing branches from one point that do not add to 1, such as 0.7 and 0.7. The second branch is what is left: 1 − 0.7 = 0.3.

Two lifts

In the application below, two lifts break down independently. The tree has the same pair of branches for lift B after each branch for lift A; part (a) multiplies along one path, and part (b) adds the two paths where exactly one lift is working.