A tree diagram is a diagram that presents multiple trials in a clear and easy-to-read manner. The branches of the tree represent the probabilities of the possible outcomes of each trial.
Theory
A tree diagram is a great way to structure different outcomes when you’re doing multiple trials. It’s also good at showing how many outcomes multiple trials have. If you have a large amount of trials, the tree diagram will become too large to give you a good overview.
Important things to remember about tree diagrams:
In the tree diagram below, you can see that ${p}_{1}$ and ${p}_{2}$ are the probabilities of the two possible outcomes of each trial. To find the probabilities of the possible outcomes of multiple trials, we have to combine these by multiplying the relevant ones together.
You can see from the tree diagram that when you have two trials that each have two possible outcomes, the possible outcomes of the multiple trials have the probabilities ${p}_{1}{p}_{1}$, ${p}_{1}{p}_{2}$, ${p}_{2}{p}_{1}$ and ${p}_{2}{p}_{2}$.
Example 1
Illustrate the different outcomes you can get by tossing a coin three times, using a tree diagram:
There are two possible outcomes for each coin toss, heads (H) and tails (T). Thus, there are
$$2\cdot 2\cdot 2=8$$ |
possible outcomes when a coin is tossed three times. This is well illustrated by means of a tree diagram:
Example 2
When giving birth, the probability of having a girl is $\text{}0.486\text{}$, while the probability of having a boy is $\text{}0.514\text{}$. The probability of the child being born colorblind is $\text{}0.044\text{}$. Make a tree diagram to show the probabilities of having a girl or boy who is or isn’t colorblind.
From the text you can see that the trials are sex and colorblindness. That gives you the following possibilities:
The trial for the sex of the baby has two possible outcomes: Boy or girl.
The trial for whether the baby is colorblind or not also has two possible outcomes: Colorblind or not colorblind.
You already know the probabilities of the baby being a boy or girl, but you still only have the probability of the baby being colorblind. That means you have to find the probability of not colorblind. Colorblind and not colorblind are complementary events, which gives you
$$P\phantom{\rule{-0.17em}{0ex}}\left(\text{notcolorblind}\right)+P\phantom{\rule{-0.17em}{0ex}}\left(\text{colorblind}\right)=1$$ |
We find the probability:
$$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(\text{notcolorblind}\right)+0.044& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P\phantom{\rule{-0.17em}{0ex}}\left(\text{notcolorblind}\right)& =1-0.044\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.956\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$If you organize this in a tree diagram, it’s pretty apparent what you need to do—namely, multiply the numbers along each branch by each other. The tree diagram ends up looking like this:
Here are the calculations: