Math 7 · 7.SP.C.8 · 7.SP.C.8B

Tree diagrams and organized lists

An experiment happens in stages, and the results need not be equally likely. How do you see every way it can turn out, and the chance of each way?

You need

Multiplying fractionsGrade 5

Unlocks

Dependent events and drawing without replacementMath 7 Finding a conditional probabilityAlgebra 2

Where you are 10 in Compound events

Probability4

Compound events10

▴ 5 above Mutually exclusive events and the addition rule The general multiplication rule

Tree diagrams and organized lists

Union and intersection of events The general addition rule

When you reach for it

Many experiments happen in stages. A customer picks an ice cream flavor and then a cone or a tub. A cyclist meets one traffic light and then another. A forecast covers Saturday and then Sunday. An outcome of the whole experiment is one result from each stage, in order, and before you can find a probability you need to see all of those outcomes, each written once.

For two stages, an organized list or a table does that. A tree diagram does the same job as a picture that keeps working for three or more stages, where a table runs out of directions. It also does a second job that a list cannot: each branch of the tree has room for the probability of its own result, so the tree still gives the right answer when the outcomes are not equally likely, which is exactly where counting outcomes stops working.

The procedure

This lesson takes two days. Part 1 draws the tree and reads it as a list of outcomes. Part 2 writes a probability on every branch and uses those numbers to find chances.

Part 1: drawing the tree and reading its paths

An ice cream stand sells three flavors, mango (M), lemon (L) and pistachio (P), each served in a cone (C) or a tub (T). An order is one flavor followed by one way of serving it.

Start at a point and draw one branch for each result of the first stage. Here that is three branches, with Mango, Lemon and Pistachio written at their ends. Between them they must cover every flavor on offer, each once, because exactly one of them is what happens at the first stage.

From the end of each branch, draw one branch for each result of the next stage. Every flavor can go in either a cone or a tub, so each of the three ends grows two branches. A third stage, such as a topping, would add a fan of branches at each of these six ends, and so on for as many stages as the experiment has.

Figure 1Every path from the starting point to a right-hand end is one order. Read from top to bottom, the ends are the organized list.

Each path is one outcome. A path is a route along the branches from the starting point to a right-hand end. Reading its labels from left to right spells the outcome: the top path is Mango then Cone, the order MC. Every order has exactly one path, since its flavor picks out one branch from the start and its serving picks out one branch from that end. Two different paths part company at some fork, so they are two different orders. That is why the ends of a tree are the sample space with nothing missed and nothing written twice.

Read the ends from top to bottom and you have the organized list.

MC, MT, LC, LT, PC, PT

This is the list made by fixing the first flavor and running through the servings, then fixing the next flavor. The tree draws that fixing as a shared branch: the two orders that start with mango share the mango branch.

Count the paths. When every fork in a stage has the same number of branches, the count is a product, here 3 \times 2 = 6: one fan of 2 for each of the 3 flavors. This is the counting principle, which counts without drawing anything. The tree is worth drawing when you need the outcomes themselves, to see which ones make up an event.

When every path is equally likely, count paths. Then a probability is a count, just as for any sample space of equally likely outcomes:

P(\text{event}) = \frac{\text{number of paths in the event}}{\text{number of paths}}

The paths are equally likely when the results at every fork are equally likely and every fork in a stage has the same number of branches. Part 2 shows why, and what to do when they are not.

Part 2: a probability on every branch

Philomena cycles to school through two crossings with traffic lights. When she arrives, the first light is green with probability 0.6 and the second with probability 0.3. The lights run on separate timers, so what the first one shows does not change the chances at the second. Call each light G (green) or R (red, which here includes amber).

The tree has two branches at each fork and four paths, but nothing says those paths are equally likely. So write the chance of each result on its branch.

The branches leaving one fork add to 1. From the start, G has 0.6, so R has 1 - 0.6 = 0.4, by the complement rule: the branches at a fork list every result of that stage, each once, and exactly one of them happens. At the second stage G has 0.3 and R has 0.7, and the same two numbers go after G and after R, because the first light does not affect the second.

Figure 2A probability on every branch. Multiplying along a path gives the chance of that path, and the four paths add to 1.

Multiply along a path for the chance of that path. Picture 100 trips to school. About 60 of them meet a green first light. The second light is green on about 30\% of trips, and since it takes no notice of the first light, on about 30\% of those 60 trips as well, which is 18 trips. So

P(\text{GG}) = 0.6 \times 0.3 = 0.18

Taking 0.3 of 0.6 is a fraction of a fraction, which is exactly what multiplying fractions computes. Every path is found the same way: P(\text{GR}) = 0.6 \times 0.7 = 0.42, P(\text{RG}) = 0.4 \times 0.3 = 0.12 and P(\text{RR}) = 0.4 \times 0.7 = 0.28.

Add the paths that make up an event. On any one trip Philomena follows exactly one path, so no two paths share an outcome, and the chance of an event made of several paths is the sum of their chances. “Exactly one green light” is GR or RG, so its probability is 0.42 + 0.12 = 0.54. “At least one green light” is every path except RR, so it is 1 - 0.28 = 0.72.

The paths add to 1. Here 0.18 + 0.42 + 0.12 + 0.28 = 1. The paths are the whole sample space, so their chances must total 1, and this is a check worth making every time. It always works when every fork adds to 1: the two paths through the first G add to 0.6 \times (0.3 + 0.7) = 0.6, the two through the first R add to 0.4, and 0.6 + 0.4 = 1.

Part 1 is the special case where every path gets the same product. Put \displaystyle \frac{1}{3} on each flavor and \displaystyle \frac{1}{2} on each serving in the ice cream tree, and every order has \displaystyle \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}, exactly what counting one path out of six gives. When the products differ, counting paths gives the wrong answer. For the lights it would say \displaystyle \frac{1}{4} for GG, but the four paths run from 0.12 to 0.42, and only the products are right.

Worked examples

Examples 1 and 2 belong to Part 1: drawing trees and counting equally likely paths. Examples 3 to 5 belong to Part 2: probabilities on the branches.

Example 1

Events in the ice cream tree

Obadiah lets chance pick his order at the stand in the procedure. He writes the six possible orders on cards, one each, and draws a card at random. (a) Find the probability that his ice cream comes in a tub. (b) Find the probability that it is not mango and comes in a cone.

Solution

The cards are the six paths MC, MT, LC, LT, PC, PT, and every card is equally likely to be drawn, so every path is equally likely.

(a) The paths ending in Tub are MT, LT and PT, one under each flavor:

P(\text{tub}) = \frac{3}{6} = \frac{1}{2}

(b) “Not mango” keeps the paths under Lemon and Pistachio, and “in a cone” keeps, of those, the ones ending in Cone: LC and PC. That is 2 paths:

P(\text{not mango, in a cone}) = \frac{2}{6} = \frac{1}{3}

Example 2

Two children

A family has two children. Use the model in which each child is equally likely to be a boy (B) or a girl (G), whatever the other child is, so that all four paths of the tree are equally likely. Find the probability that the family has one boy and one girl, and the probability that it has at least one girl.

Solution

Take the older child as the first stage and the younger as the second.

Figure 3Two children, older first. BG and GB are two different paths.

The four paths are BB, BG, GB and GG. “One boy and one girl” is BG (an older boy and a younger girl) or GB (an older girl and a younger boy). These are two different families, so they are two paths:

P(\text{one of each}) = \frac{2}{4} = \frac{1}{2}

“At least one girl” is every path with a G in it, which is BG, GB and GG:

P(\text{at least one girl}) = \frac{3}{4}

The tree is what stops BG and GB being merged into one outcome.

Example 3

A dry weekend

Magnus plans a two-day walk. The forecast gives a 20\% chance of rain on Saturday and a 25\% chance on Sunday, and he treats the two days as not affecting each other. Find the probability of (a) a dry weekend, (b) rain on exactly one day, (c) rain on at least one day.

Solution

Write the percentages as decimals. R (rain) has 0.2 on Saturday and 0.25 on Sunday, so D (dry) has 1 - 0.2 = 0.8 and 1 - 0.25 = 0.75. The tree has the shape of the lights’ tree: R and D at the first fork, and after each of them the same pair, R with 0.25 and D with 0.75.

(a) A dry weekend is the path DD: 0.8 \times 0.75 = 0.6, a 60\% chance.

(b) Rain on exactly one day is RD or DR:

\begin{aligned} 0.2 \times 0.75 + 0.8 \times 0.25 &= 0.15 + 0.2 \\ &= 0.35 \end{aligned}

which is a 35\% chance.

(c) Rain on at least one day is every path except DD, so its probability is 1 - 0.6 = 0.4, a 40\% chance. Adding the paths RR, RD and DR gives 0.05 + 0.15 + 0.2 = 0.4, the same.

Treating the days as unconnected is a choice of model, and the problem stated it. Real weather on two days running is often linked, and a forecaster who knew Saturday had been wet might change Sunday’s number. Then the branches after R and after D on Saturday would carry different numbers, which is the situation of Dependent events and drawing without replacement.

Example 4

Three penalties

In a penalty shoot-out, a team’s first three kickers each score with probability 0.7, and each kick is treated as unaffected by the others. Find the probability that (a) all three score, (b) exactly two score, (c) at least two score.

Solution

The tree has three stages, one per kick, with S (scores) on branches marked 0.7 and M (misses) on branches marked 0.3.

Figure 4Three kicks. Paths with the same number of goals have the same probability, since they multiply the same numbers in a different order.

(a) All three is the single path SSS:

0.7 \times 0.7 \times 0.7 = 0.343

(b) Exactly two goals means one miss, and the miss can be the third, second or first kick: SSM, SMS and MSS. Each of these paths multiplies two 0.7’s and one 0.3, in a different order, and the order of the factors does not change a product (the commutative property), so each has 0.7 \times 0.7 \times 0.3 = 0.147. Adding the three paths,

\begin{aligned} 0.147 + 0.147 + 0.147 &= 3 \times 0.147 \\ &= 0.441 \end{aligned}

The goals column in the figure groups the eight paths as 1, 3, 3 and 1. As a check, the remaining paths, one goal (3 \times 0.063 = 0.189) and no goals (0.3 \times 0.3 \times 0.3 = 0.027), bring the total to 0.343 + 0.441 + 0.189 + 0.027 = 1.

Example 5

Kicking until a goal

In practice, Rupert keeps taking penalties until he scores, but stops after three kicks whatever happens. He scores each kick with probability 0.6, unaffected by his earlier kicks. Find the probability that (a) he scores, (b) he takes all three kicks.

Solution

The tree is not full this time. After a goal Rupert stops, so a branch ending in S has no fork after it; only a miss leads to another kick.

Figure 5A tree need not be full. After a goal the path stops, so the four paths have three different lengths.

There are four paths, of three different lengths:

Path	Along the path	Probability
S	0.6	0.6
MS	0.4 \times 0.6	0.24
MMS	0.4 \times 0.4 \times 0.6	0.096
MMM	0.4 \times 0.4 \times 0.4	0.064

They add to 0.6 + 0.24 + 0.096 + 0.064 = 1, so no outcome is missing.

(a) He scores on the paths S, MS and MMS: 0.6 + 0.24 + 0.096 = 0.936. The only other path is MMM, so this is also 1 - 0.064.

(b) He takes a third kick exactly when he misses the first two, which is the paths MMS and MMM: 0.096 + 0.064 = 0.16. That equals 0.4 \times 0.4, the chance of the two misses, because the two branches after the second miss add to 1.

The organized list of this experiment has four entries, not the 2 \times 2 \times 2 = 8 that three full stages would give.

Boundaries

The branches at a fork must list every result of that stage, each once. Leave one out and some outcomes have no path; list two that overlap and some outcome has two paths. Either way the probabilities at that fork no longer add to 1, which is how the slip shows.

Counting paths gives probabilities only when every path has the same chance. That is sure to hold when the results at every fork are equally likely and every fork in a stage has the same number of branches. Philomena’s lights break the first condition, and a menu where one flavor comes two ways and another three ways breaks the second (Exercise 17). Multiplying along the branches works in both cases.

Multiplying along a path needs the right number on each branch. In this lesson the stages do not affect each other, so a branch carries the plain chance of its result. When an earlier result changes the chance of a later one, as when two counters are drawn from a bag without putting the first back, the branch must carry the chance after that change, and the plain chance gives a wrong product. That case belongs to Dependent events and drawing without replacement and the general multiplication rule.

Add paths, not overlapping events. Paths never overlap, so their chances add. Two events can overlap: “rain on Saturday” and “rain on Sunday” in Example 3 share the path RR, and adding 0.2 + 0.25 = 0.45 counts RR twice. The right answer, 0.4, is less by exactly 0.05, the chance of RR. Adding events that overlap is the general addition rule.

A tree grows fast. Two branches at each of ten stages give 2^{10} = 1024 paths, too many to draw. A tree is for experiments with a few stages. For more, count with the counting principle, and find “at least one” chances through the complement, as in At least one problems.

Where this goes next

The tree is the working picture for every compound event in this part of the map. When one stage changes the next, the same tree carries updated numbers on its later branches, which is drawing without replacement, and multiplying along a path becomes the general multiplication rule. When the stages do not affect each other, the products are the multiplication rule for independent events.

Read backwards, a tree answers a new kind of question: given that Philomena met exactly one green light, how likely is it that it was the first one? That is conditional probability, and its formula divides the chance of one path by the chance of the whole event.

Exercises

Part 1: drawing trees and counting paths

A smoothie bar blends banana (B) or kiwi (K) with one base: oat milk (o), apple juice (a) or coconut water (c). Draw the tree, list every smoothie it shows in the order of its ends, and count them.
A smoothie from Exercise 1 is chosen at random, each equally likely. Find the probability that it is (a) kiwi without oat milk, (b) made with oat milk or coconut water.
For a family with two children, each child equally likely to be a boy or a girl whatever the other is, find the probability that (a) both are girls, (b) the older child is a boy.
Using the same model for three children, draw the tree and find the probability that (a) there is exactly one boy, (b) the youngest is the only girl.
Add a size, small (s) or large (l), as a third stage after the base in Exercise 1. How many paths does the tree have? If a smoothie is chosen at random, each path equally likely, what is the probability that it is banana with coconut water?
A tree has 4 branches from its starting point, 3 from the end of each of those, and 2 from the end of each of those. How many paths does it have, and how many of them pass through the top branch at the start?
On Fridays banana comes only with oat milk or apple juice, while kiwi comes with all three bases. Draw the tree and list the smoothies. Why is the count not 2 \times 3?

Part 1: reasoning

Someone says a family with two children can have two boys, two girls, or one of each, so the chance of one of each is \displaystyle \frac{1}{3}. Use the tree to explain what is wrong.

Part 2: probabilities on the branches

On another route, the first light is green with probability 0.5 and the second with probability 0.7, neither affecting the other. Find the probability that (a) both are green, (b) neither is green, (c) exactly one is green.
Find the missing branch probability at each fork: (a) two branches, one marked 0.35; (b) three branches, marked 0.2, 0.45 and one blank.
The forecast gives a 10\% chance of rain on Saturday and 30\% on Sunday, the days treated as not affecting each other. Find the probability of (a) a dry weekend, (b) rain on exactly one day, (c) rain on at least one day.
Nerissa takes two penalties and scores each with probability \displaystyle \frac{3}{4}, the kicks not affecting each other. Find the probability that she scores (a) twice, (b) exactly once, (c) not at all.
At the ice cream stand of the procedure, customers choose mango with probability 0.5, lemon with 0.3 and pistachio with 0.2, and then a cone with probability 0.4 and a tub with 0.6, whatever the flavor. Find the probability that a customer has (a) lemon in a tub, (b) a cone, by adding paths.
Three kickers each score with probability 0.9, each kick unaffected by the others. Find the probability that (a) all three score, (b) exactly two score, (c) at most one scores.
On Philomena’s route the first light is green with probability 0.4, and the chance that both are green is 0.1. The lights do not affect each other. What is the chance that the second light is green?
Rupert kicks until he scores, but at most twice, and scores each kick with probability 0.3. List the paths with their probabilities, and find the probability that he scores.
Use the Friday menu of Exercise 7. A customer picks the fruit at random, then a base at random from those offered with it. Find the probability of each smoothie, and the probability that it is made with oat milk.
Leopold scores a penalty with probability 0.85 and Seraphina with probability 0.6, and neither kick affects the other. Leopold kicks first. Find the probability that (a) exactly one of them scores, (b) at least one scores.

Part 2: reasoning

Explain why the probabilities on the branches leaving any one fork add to 1.
Using 100 trips through the lights of Exercise 9, explain why the chance of the path “green, then green” is 0.5 \times 0.7.
Explain why the probabilities of all the paths of a two-stage tree add to 1, whatever numbers are on the branches, as long as each fork adds to 1.
Obadiah says that the chance of two green lights on Philomena’s route in the procedure is \displaystyle \frac{1}{4}, because GG is one path out of four. Explain what is wrong, and say when counting paths does give the probability.
In Example 4, explain why the paths SSM, SMS and MSS all have the same probability, and why the chance of exactly two goals is that probability times 3.

Answers

6 smoothies: Bo, Ba, Bc, Ko, Ka, Kc. Two branches from the start and three from each end, so 2 \times 3 = 6.
(a) \displaystyle \frac{1}{3}: Ka and Kc, \displaystyle \frac{2}{6}. (b) \displaystyle \frac{2}{3}: Bo, Bc, Ko and Kc, \displaystyle \frac{4}{6}.
(a) \displaystyle \frac{1}{4}: only GG. (b) \displaystyle \frac{1}{2}: BB and BG, \displaystyle \frac{2}{4}.
The tree has 2 \times 2 \times 2 = 8 paths. (a) \displaystyle \frac{3}{8}: BGG, GBG and GGB. (b) \displaystyle \frac{1}{8}: only BBG.
12 paths, 2 \times 3 \times 2. Banana with coconut water is Bcs or Bcl, so the probability is \displaystyle \frac{2}{12} = \frac{1}{6}.
24 paths, 4 \times 3 \times 2, and 6 of them pass through the top branch, 3 \times 2.
5 smoothies: Bo, Ba, Ko, Ka, Kc. The two forks after the fruits have different numbers of branches, 2 and 3, so the count is 2 + 3 = 5. The product needs every fork in a stage to have the same number of branches.
“One of each” is two paths, BG and GB: an older boy with a younger girl, and an older girl with a younger boy. The four paths BB, BG, GB and GG are equally likely, so the three descriptions are not. One of each has chance \displaystyle \frac{2}{4} = \frac{1}{2}, and two boys and two girls have \displaystyle \frac{1}{4} each.
(a) 0.35: 0.5 \times 0.7. (b) 0.15: 0.5 \times 0.3. (c) 0.5: GR or RG, 0.15 + 0.35.
(a) 0.65: 1 - 0.35. (b) 0.35: 1 - 0.2 - 0.45.
(a) 0.63: 0.9 \times 0.7. (b) 0.34: 0.1 \times 0.7 + 0.9 \times 0.3 = 0.07 + 0.27. (c) 0.37: 1 - 0.63.
(a) \displaystyle \frac{9}{16}: \displaystyle \frac{3}{4} \times \frac{3}{4}. (b) \displaystyle \frac{3}{8}: the paths SM and MS, \displaystyle \frac{3}{16} + \frac{3}{16} = \frac{6}{16}. (c) \displaystyle \frac{1}{16}: \displaystyle \frac{1}{4} \times \frac{1}{4}.
(a) 0.18: 0.3 \times 0.6. (b) 0.4: the paths MC, LC and PC have 0.5 \times 0.4, 0.3 \times 0.4 and 0.2 \times 0.4, which add to 0.2 + 0.12 + 0.08. That equals the cone branch, 0.4, because the three flavor branches add to 1.
(a) 0.729: 0.9 \times 0.9 \times 0.9. (b) 0.243: three paths, each 0.9 \times 0.9 \times 0.1 = 0.081. (c) 0.028: 1 - 0.729 - 0.243, or no goals, 0.001, plus exactly one goal, 3 \times 0.009 = 0.027.
0.25. The path GG has probability 0.4 \times p = 0.1, so p = 0.1 \div 0.4.
S with 0.3, MS with 0.7 \times 0.3 = 0.21, MM with 0.7 \times 0.7 = 0.49, which add to 1. He scores with probability 0.3 + 0.21 = 0.51, which is also 1 - 0.49.
Bo and Ba have \displaystyle \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} each; Ko, Ka and Kc have \displaystyle \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} each. Oat milk is Bo or Ko: \displaystyle \frac{1}{4} + \frac{1}{6} = \frac{5}{12}, not \displaystyle \frac{2}{5}, since the five smoothies are not equally likely.
(a) 0.43: 0.85 \times 0.4 + 0.15 \times 0.6 = 0.34 + 0.09. (b) 0.94: everything except both missing, 1 - 0.15 \times 0.4.
The branches at a fork are all the results that stage can have, each listed once, so exactly one of them happens. Their chances therefore add to the chance that the stage has some result, which is 1; each branch is the complement of all the others together.
The first light is green on about 50 of the 100 trips. The second light is green on about 70\% of all trips, and since it does not depend on the first, on about 70\% of those 50 as well, which is 35 trips. So the path has chance \displaystyle \frac{35}{100} = 0.5 \times 0.7: a fraction of a fraction.
Group the paths by their first branch. If a first branch has probability a and the branches at the fork after it have q_1, q_2, \ldots, its paths have a q_1, a q_2, \ldots, which add to a(q_1 + q_2 + \cdots) = a \times 1 = a. Adding these over all the first branches gives the sum of the numbers at the first fork, which is 1.
The four paths are not equally likely: GG has 0.18 while GR has 0.42. Counting works when every path has the same probability, which is sure to happen when the results at every fork are equally likely and every fork in a stage has the same number of branches.
Each of the three paths multiplies two 0.7’s and one 0.3, only in a different order, and a product does not depend on the order of its factors, so each is 0.147. They are different paths, so no two happen together, and the chance that one of them happens is their sum, 0.147 + 0.147 + 0.147 = 3 \times 0.147.