Introduction to Statistics · Chapter 3 · Probability Topics

Tree and Venn Diagrams

Two pictures for the same probabilities: a tree lays out every path of a staged experiment, and a Venn diagram turns AND, OR, and neither into regions you can point at.

bookSHelf · Introduction to Statistics · §3.5 · a self-paced section

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Learning objectives — by the end of this section you will be able to

Objectives

Build a tree diagram for a two-stage experiment, branches labeled with frequencies or probabilities §3.5.1
Read a probability off a tree by multiplying along one path and adding across paths §3.5.1
Explain how with-replacement and without-replacement sampling change the second set of branches §3.5.1
Draw a Venn diagram for two events and shade AND, OR, and neither §3.5.2
Pull $P(A \text{ AND } B)$ , $P(A \text{ OR } B)$ , and $P(A \mid B)$ off a completed diagram §3.5.2

3.5

§3.5.1 — every path a staged experiment could take

A tree diagram lays out a multi-stage experiment stage by stage: one set of branches per stage, one path from trunk to tip per outcome. Two rules run the whole section.

Multiply along a path. Add across paths.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

§3.5.1 — the picture behind every staged experiment

Definition 3.5.1: Tree Diagram

Definition 3.5.1 — Tree Diagram

A tree diagram is a graph that shows every possible outcome of a multi-stage experiment. Each stage gets its own set of branches, and each branch is labeled with either the frequency or the probability of that step. A complete route from the trunk to a tip is called a path, and each path is one distinct outcome of the whole experiment.

Multiply along a path. The probability of one complete outcome is the product of the probabilities on its branches.
Add across paths. The probability of an event that several paths satisfy is the sum of those paths' probabilities.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

§3.5.1 — the pool resets after every draw

Definition 3.5.2: Sampling With Replacement

Definition 3.5.2 — Sampling With Replacement

You sample with replacement when the object selected is returned to the pool before the next selection is made. The pool is identical at every stage, so the second set of branches carries the same probabilities as the first, and the stages are independent.

Def. 3.5.1: multiply along one path, add across several.

Def. 3.5.2: the ball goes back, so draw 2 faces the same urn.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Insight Note — a tree is a map of every way the day could go

Each fork is a moment the story could break either way.

Each tip is one complete story from start to finish. Multiplying along a path asks “how likely is this exact story?” Adding paths asks “how likely is any story that ends this way?”

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

§3.5.1 — the pool shrinks after every draw

Definition 3.5.3: Sampling Without Replacement

Definition 3.5.3 — Sampling Without Replacement

You sample without replacement when the object selected is kept out of the pool. The pool shrinks by one at every stage, so the second set of branches depends on which branch you took first, and the stages are dependent.

Definition 3.5.3: the ball stays out, so draw 2 faces a smaller urn.

In the with-replacement case every second-stage fork looks the same, so a formula would do. In the without-replacement case each second-stage fork carries its own numbers — the tree does that bookkeeping for you.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — frequencies through a tree, with replacement

Try It Now 3.5.1

Try It Now 3.5.1 — a standard deck, two draws with replacement

In a standard deck of 52 cards, 12 are face cards ( $F$ ) and 40 are not ( $N$ ). Draw two cards, one at a time, with replacement. Using a tree diagram of frequencies, find $P(FF)$ .

$P(FF) = \frac{144}{2{,}704} = \frac{9}{169} \approx 0.0533$

144 is the FF path's frequency out of $52(52)=2{,}704$ total outcomes.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — two draws, with replacement

Example 3.5.1 · Two Draws From an Urn, With Replacement

First draw	Second draw	Outcome	Number of ways
R (3)	R (3)	RR	9
R (3)	B (8)	RB	24
B (8)	R (3)	BR	24
B (8)	B (8)	BB	64

Table 3.5.2: every path for two draws with replacement, as frequencies out of 121 — an urn of 3 red and 8 blue, $11(11)=121$ .

b) $P(RR)$ c) $P(RB \text{ OR } BR)$ d) $P(R \text{ on 1st AND } B \text{ on 2nd})$ e) $P(R \text{ on 2nd} \mid B \text{ on 1st})$ f) $P(BB)$ g) $P(B \text{ on 2nd} \mid R \text{ on 1st})$

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Example 3.5.1 — one tree, six different questions

Reading all six off Table 3.5.2

Part	Question	Answer
b	$P(RR)$	$\frac{9}{121}$
c	$P(RB \text{ OR } BR)$	$\frac{48}{121}$
d	$P(R \text{ 1st AND } B \text{ 2nd})$	$\frac{24}{121}$
e	$P(R \text{ 2nd} \mid B \text{ 1st})$	$\frac{3}{11}$
f	$P(BB)$	$\frac{64}{121}$
g	$P(B \text{ 2nd} \mid R \text{ 1st})$	$\frac{8}{11}$

Parts e and g both equal the plain probability of that color on a single draw — that is what replacement buys you: the two draws are independent.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — one of each color, without replacement

Try It Now 3.5.2

Try It Now 3.5.2 — four red, three yellow, no replacement

A box holds four red balls and three yellow balls. Two are drawn without replacement. Find the probability that one ball of each coloring is selected.

$P(RY)+P(YR)=\left(\frac{4}{7}\right)\left(\frac{3}{6}\right)+\left(\frac{3}{7}\right)\left(\frac{4}{6}\right)=\frac{24}{42}=\frac{4}{7}\approx0.571$

Two different paths give “one of each”, so add them — every second-stage denominator is 6 because the first ball stayed out.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — the same urn, without replacement

Example 3.5.2 · The Same Urn, Without Replacement

First draw	Second draw	Outcome	Path probability
$R$ : $\frac{3}{11}$	$R$ : $\frac{2}{10}$	RR	$\frac{6}{110}$
$R$ : $\frac{3}{11}$	$B$ : $\frac{8}{10}$	RB	$\frac{24}{110}$
$B$ : $\frac{8}{11}$	$R$ : $\frac{3}{10}$	BR	$\frac{24}{110}$
$B$ : $\frac{8}{11}$	$B$ : $\frac{7}{10}$	BB	$\frac{56}{110}$

Table 3.5.3: the same 3-red, 8-blue urn — every second-stage denominator drops from 11 to 10 because the first marble stays out.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Example 3.5.2 — the urn that remembers its first draw

Six probabilities, this time dependent

Part	Question	Answer
a	$P(RR)$	$\frac{6}{110}$
c	$P(R \text{ 2nd} \mid B \text{ 1st})$	$\frac{3}{10}$
e	$P(BB)$	$\frac{56}{110}$
f	$P(B \text{ 2nd} \mid R \text{ 1st})$	$\frac{8}{10}=\frac{4}{5}$

Compare part f with Example 3.5.1's part g: with replacement it was $\frac{8}{11}$ ; without replacement it is $\frac{8}{10}$ . Removing a red marble raised the chance the next one is blue — that is exactly what “dependent” means.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Context Pause — why the second denominator is 10 and not 11

You drew a marble and did not put it back.

There are ten marbles in the urn the second time you reach in. Every second-stage branch has to be measured against that smaller urn — forgetting to shrink the denominator is the single most common mistake in without-replacement problems.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Try it in rāSHio

Make the tool compute the wrong urn on purpose

Open Calculators → Probability, switch to Conditional, and enter $P(A)=0.2727$ and $P(B)=0.2727$ — both $\frac{3}{11}$ , the chance of drawing red — with Assume independence ticked.

The panel returns $P(A \text{ AND } B) = 0.07438 = \frac{9}{121}$ — Example 3.5.1's with-replacement answer, not this one. The tree above gives $\frac{6}{110} = 0.0545$ instead; the gap between them is the whole content of Definition 3.5.3 — the panel has no way to know you kept the marble out, so you have to supply the shrunken second rate yourself.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Figure 3.5.1 — Calculators → Probability, Conditional mode

Watching rāSHio assume independence

Figure 3.5.1: with Assume independence ticked, rāSHio returns the AND as the product of two unchanged rates — the with-replacement answer.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — four tabby, five black, no replacement

Example 3.5.3 · Adopting Two Kittens

First kitten	Second kitten	Outcome	Path probability
$T$ : $\frac{4}{9}$	$T$ : $\frac{3}{8}$	TT	$\frac{12}{72}$
$T$ : $\frac{4}{9}$	$K$ : $\frac{5}{8}$	TK	$\frac{20}{72}$
$K$ : $\frac{5}{9}$	$T$ : $\frac{4}{8}$	KT	$\frac{20}{72}$
$K$ : $\frac{5}{9}$	$K$ : $\frac{4}{8}$	KK	$\frac{20}{72}$

Table 3.5.4: nine kittens, four tabby and five black; every second-stage denominator is 8.

a) both tabby? b) one of each coloring? c) tabby second, given black first? d) two of the same color?

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Example 3.5.3 — read the tree, don't guess a formula

The trap hiding in part a

Part	Answer
a) both tabby	$\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)$
b) one of each	$\frac{20}{72}+\frac{20}{72}=\frac{40}{72}$
c) tabby 2nd $\mid$ black 1st	$\frac{4}{8}=\frac{1}{2}$
d) same color	$\frac{12}{72}+\frac{20}{72}=\frac{32}{72}=\frac{4}{9}$

Two of the four multiple-choice options for part a ignore the shrinking litter entirely — $\left(\frac{1}{2}\right)^2$ and $\left(\frac{4}{9}\right)^2$ both assume the first kitten came back. The tree is what keeps that bookkeeping honest.

3.5

§3.5.2 — every region you can point at

A Venn diagram is a box for the sample space with circles inside it for the events. Once it is drawn, AND, OR, and neither are three regions you can point to.

The box is what makes probabilities add to 1.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

§3.5.2 — a picture of an experiment's outcomes

Definition 3.5.4: Venn Diagram

Definition 3.5.4 — Venn Diagram

A Venn diagram is a picture of the outcomes of an experiment: a box for the sample space $S$ , with circles or ovals inside it for the events. An outcome sits inside a circle exactly when it belongs to that event.

$A \text{ AND } B$ is the overlap where the two circles cross.
$A \text{ OR } B$ is everything inside either circle, overlap counted once.
Inside the box but outside both circles: neither event.

Definition 3.5.4: AND, OR, and neither are three regions you can point at.

A Venn diagram is not just two circles; it is two circles inside a box, and the box is what makes probabilities add to 1 — the size of the leftover “neither” region is often the thing a question is really asking about.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Insight Note — the box is the whole world, and the circles are clubs inside it

Standing in the overlap means you belong to both clubs.

Standing in the box but outside every circle means you joined nothing — and those people still count, which is why the rectangle is drawn at all.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — place eight outcomes in two circles

Try It Now 3.5.3

Try It Now 3.5.3 — colors, equally likely

Outcomes black, white, red, orange, yellow, green, blue, purple, each equally likely. Let $C=\{\text{green, blue, purple}\}$ and $P=\{\text{red, yellow, blue}\}$ . Draw a Venn diagram representing this situation.

Overlap ( $C$ and $P$ ): blue. $C$ only: green, purple. $P$ only: red, yellow. Outside both: black, white, orange — $1+2+2+3=8$ , the whole sample space.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — reading AND and OR straight off the picture

Example 3.5.4 · Twelve Equally Likely Outcomes

Worked Example 3.5.4 — outcomes 1 through 12

Outcomes 1, 2, …, 12, each equally likely. $A=\{1,2,3,4,5,6\}$ and $B=\{6,7,8,9\}$ . Find $A \text{ AND } B$ and $A \text{ OR } B$ .

$A \text{ AND } B = \{6\} \qquad A \text{ OR } B = \{1,2,3,4,5,6,7,8,9\}$

$P(A \text{ OR } B)=\frac{9}{12}=0.75 \qquad P(A \text{ AND } B)=\frac{1}{12}\approx0.083$

Check: $\frac{6}{12}+\frac{4}{12}-\frac{1}{12}=\frac{9}{12}$ , which agrees.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — a single die, two events

Try It Now 3.5.4

Try It Now 3.5.4 — prime and odd

Roll a fair six-sided die. $A$ = a prime number of dots is rolled, $B$ = an odd number of dots is rolled. $S=\{1,2,3,4,5,6\}$ . Draw a Venn diagram representing this situation.

Overlap (prime and odd): 3, 5. $A$ only: 2. $B$ only: 1. Outside both: 4, 6.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — two fair coins, four outcomes

Example 3.5.5 · Flipping Two Fair Coins

Worked Example 3.5.5 — two coins

$A$ = tails on the first coin, $B$ = tails on the second coin. Identify $A \text{ AND } B$ , $A \text{ OR } B$ , and any outcome that belongs to neither event.

$S=\{TT,TH,HT,HH\} \qquad A \text{ AND } B=\{TT\} \qquad A \text{ OR } B=\{TT,TH,HT\}$

HH sits inside the box but outside both circles — it has no tails at all, so it belongs to neither event.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — two circles that overlap by only 5%

Try It Now 3.5.5

Try It Now 3.5.5 — second jobs and working spouses

50% of factory workers work a second job ( $J$ ), 25% have a spouse who also works ( $W$ ), and 5% do both. Draw a Venn diagram showing the relationships.

$P(J \text{ only})=0.45 \qquad P(W \text{ only})=0.20 \qquad P(\text{neither})=1-(0.45+0.05+0.20)=0.30$

Peel the 5% overlap out of each circle first, then let the whole box add to 1.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — build the diagram region by region

Example 3.5.6 · Clubs and Part-Time Work

Worked Example 3.5.6 — 40% club, 50% part time, 5% both

40% of students belong to a club ( $C$ ), 50% work part time ( $PT$ ), 5% do both. Find a) $P(C)$ b) $P(PT)$ c) $P(C \text{ AND } PT)$ d) $P(C \mid PT)$ e) $P(C \text{ OR } PT)$ .

Start from the overlap and work outward: overlap 0.05, $C$ only $0.40-0.05=0.35$ , $PT$ only $0.50-0.05=0.45$ , neither $1-(0.35+0.05+0.45)=0.15$ .

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Example 3.5.6 — five reads off one diagram

The conditional shrinks the world to one circle

Part	Answer
a) $P(C)$	0.40
b) $P(PT)$	0.50
c) $P(C \text{ AND } PT)$	0.05
d) $P(C \mid PT)$	$\frac{0.05}{0.50}=0.1$
e) $P(C \text{ OR } PT)$	$0.40+0.50-0.05=0.85$

That matches the diagram directly: $0.35+0.05+0.45=0.85$ , which is also $1-0.15$ .

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Your turn — add an oval for a third event

Try It Now 3.5.6

Try It Now 3.5.6 — novels and non-fiction

$P(\text{novel})=0.6$ , $P(\text{non-fiction})=0.4$ , $P(\text{both})=0.2$ . b) Find $P(\text{novel OR non-fiction})$ . d) Some customers buy only compact disks — where does a new oval for that event go?

$P(N \text{ OR } F)=0.6+0.4-0.2=0.8$

A new oval for CD-only buyers sits inside the box, disjoint from both circles — it belongs to the $1-0.8=0.2$ “neither” region.

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Worked example — a rectangle, a circle, and an oval

Example 3.5.7 · Blood Type and Rh Factor

Worked Example 3.5.7 — universal donors

Among African Americans: 4% have type O blood and Rh− ( $O$ AND $R$ ), 7.5% have Rh− (the midpoint of 5–10%, $R$ ), and 51% have type O ( $O$ ). Find $P(O)$ , $P(R)$ , $P(O \text{ AND } R)$ , and $P(O \text{ OR } R)$ .

$P(O)=0.51 \qquad P(R)=0.075 \qquad P(O \text{ AND } R)=0.04$

$P(O \text{ OR } R)=0.51+0.075-0.04=0.545$

The overlap is the universal donors this example opened with; outside both shapes is neither, probability $1-0.545=0.455$ .

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

Key Terminology — the six terms this section defined

The vocabulary

Trees

tree diagram — a graph of a multi-stage experiment; each stage is a set of branches, each complete path one outcome.
branch — a single labeled edge, carrying the frequency or probability of one step.
path — a complete route from trunk to tip; its probability is the product of its branches.

Sampling and pictures

with replacement — the pool and its probabilities are unchanged, and the stages are independent.
without replacement — the pool shrinks, and the second-stage probabilities depend on the first.
Venn diagram — a box for the sample space $S$ with circles or ovals inside it for the events.

3.5

The headline result of §3.5

Two pictures, the same probabilities

A tree counts paths — multiply along one, add across several. A Venn diagram counts regions — AND is the overlap, OR is everything inside either circle. Both read a probability straight off a picture.

$P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B)$

† Whether the pool resets (with replacement, independent stages) or shrinks (without replacement, dependent stages) changes only the second set of tree branches — the box-and-circles picture of a Venn diagram never depends on how the outcomes were generated.

3.5

Tree and Venn Diagrams · bookSHelf Intro Stats§3.5

§3.5 — conclusions

What §3.5 leaves you with

The core idea

A tree lays out every path of a staged experiment — multiply along one path, add across several that share an outcome. A Venn diagram lays out every region of two events at once — AND is the overlap, OR is everything inside either circle, and outside both circles still counts.

The failure case

Forgetting that without-replacement shrinks the second-stage denominator, or forgetting that a Venn diagram's box has to leave room for “neither” — both are the same mistake: dropping part of the sample space.

Next: §3.6 — Probability Topics, the chapter review.