The tree diagram

A tree diagram shows a two-stage experiment stage by stage, with each outcome’s probability beside it. It works when the outcomes are not equally likely too: then the cells of an outcome table are not equally likely either, and counting them is not enough.

The experiment in the examples:
A spinner has five equal parts: three red and two blue.
It is spun twice.
In each spin, the probability of red is $\displaystyle \frac{3}{5}$, and of blue $\displaystyle \frac{2}{5}$.

The example below builds the tree diagram of the two spins, a stage at a time:

• A branch is a line leading to one outcome of a stage, with that outcome’s probability on it.
• The first stage has two branches: red, $\displaystyle \frac{3}{5}$, and blue, $\displaystyle \frac{2}{5}$. Together they make 1.
• From the end of each branch come the branches of the second stage – $\displaystyle \frac{3}{5}$ and $\displaystyle \frac{2}{5}$ again, since the spinner has not changed.
• A path goes from the start to the end, one branch in each stage. Each path is one outcome of the experiment, red then blue for example. This tree has 4 paths.
• Each path’s probability is written at its end.

A path’s probability is the product of the probabilities on its branches.

Why multiply: say the experiment – two spins – is done 100 times.

• In about $\displaystyle \frac{3}{5}$ of them, 60 times, the first spin is red.
• Of those 60 times, in about $\displaystyle \frac{3}{5}$, 36 times, the second spin is red too.
• 36 times out of 100 is $\displaystyle \frac{3}{5}$ of $\displaystyle \frac{3}{5}$:

When an event has several paths, their probabilities are added. The example below finds the probability that both spins are the same colour. The event’s paths are marked:

• They are added because each time the experiment is done, only one path comes out.
• The probabilities of all the paths add up to 1.

When all the outcomes are equally likely, the tree diagram and the outcome table give the same answer. In the tree below a coin is tossed and then a die is thrown, and the paths of the event “heads and an even number” are marked:

• Each path’s probability: $\displaystyle {\frac{1}{2}\cdot\frac{1}{6}}={\frac{1}{12}}$ – the same as each of the 12 cells in the outcome table of a coin and a die.
• The event has 3 paths: $\displaystyle {\frac{1}{12}+\frac{1}{12}+\frac{1}{12}}={\frac{3}{12}}={\frac{1}{4}}$.

The probability of “at least once” is easiest found through the complement, “never”: $\displaystyle {P(\overline{A})}={1-P(A)}$. The example below finds the probability that at least one of the two spins is blue.

• The complement: neither spin is blue, so both are red. That is a single path.
• Its probability: $\displaystyle {\frac{3}{5}\cdot\frac{3}{5}}={\frac{9}{25}}$.
• The probability of blue at least once: $\displaystyle {1-\frac{9}{25}}={\frac{16}{25}}$.
• Without the complement, three paths are added: $\displaystyle {\frac{6}{25}+\frac{6}{25}+\frac{4}{25}}={\frac{16}{25}}$. The more stages, the longer that way is.

A quarter of a spinner is yellow, and the rest green.
It is spun twice.
Write the probabilities on the branches and each path’s probability.
Mark the paths of the event “both spins are the same colour”, and find its probability.

A coin is tossed, then a spinner with three equal parts, 1, 2, 3, is spun.
Write the probabilities on the branches and each path’s probability.
Mark the paths of the event “the coin shows heads, or the spinner shows 3”, and find its probability.

Noa scores a free throw with probability $\displaystyle 0.7$.
She takes two free throws.
Write the probabilities on the branches and each path’s probability.
Find the probability that Noa scores at least once.

A coin is tossed three times.
Write the probabilities on the branches and each path’s probability.
Mark the path of the event “heads never comes up”.
Find the probability of heads at least once.