The last two displays of the chapter, for when a probability problem is complex enough to be worth drawing. A tree diagram lays out a sequence of stages with a probability or a frequency on every branch, so a path's probability is the product along it and a conditional probability is read by keeping one subtree and discarding the rest — which makes it the multiplication rule of section 3.3 drawn, with the conditional impossible to overlook. A Venn diagram shows two events as overlapping regions with a count in each, so unions, intersections and complements all become sums of regions. Closes by choosing among the tree, the Venn diagram and the contingency table.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 3 — Probability Topics
Tree and Venn Diagrams
Objectives
Five outcomes. The second and third are the reasons the tree is worth drawing at all.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 189-195 — the section these objectives are drawn from
Warm-up
Section 3.3's multiplication rule says P(A AND B) is P(A) times P(B given A), and section 3.2 said drawing without replacement makes the two draws dependent.
Discussion prompt
An urn holds three red balls and eight blue. Two are drawn without replacement. Write down the probability of red then blue, then of blue then red, and notice something about the two answers.
Hint: Work each one with the multiplication rule, updating the urn after the first draw.
Answer:
Red then blue is three elevenths times eight tenths, which is 24 over 110. Blue then red is eight elevenths times three tenths, which is also 24 over 110.
The two orders have the same probability, even though the individual branch probabilities are quite different — the numerators and denominators simply appear in a different order. That is not a coincidence but it is not obvious either, and it is easy to miss when the calculations are done separately.
Laid out as a tree, both paths are visible at once and the coincidence is apparent. That is what the displays in this section are for: the book says plainly that when probability problems are complex it can be helpful to graph the situation.
Concept
When probability problems are complex it can be helpful to graph the situation. A tree diagram is a special type of graph used to determine the outcomes of an experiment, consisting of branches labelled with either frequencies or probabilities. A Venn diagram shows events as regions, so their overlaps can be read directly.
tree diagram — A graph whose branches represent the stages of an experiment, each labelled with a frequency or a probability. The outcomes of the experiment are the paths from the root to the leaves, and a path's probability is the product of the branch probabilities along it.
\[ P(\text{path}) = \text{product of the branch probabilities along it} \]
The two displays suit different shapes of problem, and neither replaces section 3.4's contingency table. A tree is for a SEQUENCE — draw one ball, then another — and its branches naturally carry conditional probabilities. A Venn diagram is for two properties of a SINGLE trial, and its regions naturally carry joint counts. A contingency table does the same job as a Venn diagram for two variables and is easier to read when both have several categories.
Figure (svg): Two columns comparing the tree diagram, which suits a sequence of stages, with the Venn diagram, which suits overlapping events
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 189
Section
Section 1
Concept
A tree diagram consists of branches labelled with either frequencies or probabilities. With frequencies, the first set of branches represents the first stage of the experiment and the second set represents the second, and the number at each leaf is the count of outcomes following that path.
branches and paths — The first set of branches represents the first draw and the second set the second draw. A path from the root to a leaf is one kind of outcome, and the leaf's number is how many of the sample space's outcomes follow that path.
\[ \text{total outcomes} = 11 \times 11 = 121 \]
The book's urn holds eleven balls, three red and eight blue, drawn twice with replacement. Because the urn is restored, both draws have eleven possibilities and the sample space has 121 outcomes. The book is careful to say why the leaf counts are what they are: the nine RR outcomes can be listed individually as R1R1, R1R2 and so on through R3R3, and the 24 BR outcomes similarly — every ball is distinguishable, which is what makes the 121 outcomes equally likely and the counting valid.
Figure (svg): A tree diagram for two draws from an urn with replacement, with frequencies on the branches and the twenty-one outcome counts at the leaves
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 189-190 — the frequency tree for the urn with replacement
Picture it
Example 3.24, with counts on the branches.
Figure (svg): A tree diagram for two draws from an urn with replacement, with frequencies on the branches and the twenty-one outcome counts at the leaves
The four leaf counts are 9, 24, 24 and 64, and they total 121 — the whole sample space, partitioned by which path was taken. That total is the tree's own check, exactly like the margin check on a contingency table: if the leaves do not sum to the size of the sample space, a branch has been miscounted.
Worked example
Example 3.24b. With frequencies, the answer is a leaf over the total.
\[ P(RR) \text{ for two draws with replacement from } 3 \text{ red and } 8 \text{ blue} \]
Count the outcomes on the RR path
Why: Three red choices, then three again.
\[ 3 x 3 = 9 \]
Count the whole sample space
Why: Eleven choices, then eleven.
\[ 11 x 11 = 121 \]
Divide
Why: Nine of one hundred twenty-one.
\[ \frac{9}{121} \]
Check against the other leaves
Why: Nine plus 24 plus 24 plus 64.
\[ 121,\text{ as required} \]
Figure (svg): The solution to Worked example a probability from a frequency tree shown as a ladder of expressions, one row per legal move
\[ P(RR) = \frac{9}{121} \]
Verify: confirm the leaves partition the sample space
Why: The four leaf counts total exactly 121, so every one of the 121 outcomes lies on exactly one path and none is counted twice. That is what makes each probability a simple leaf over total, with no addition rule needed — the paths are mutually exclusive by construction. It is also the check that catches an arithmetic slip in any branch, since one wrong count breaks the total.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 190
Faded example
Eleven balls, three red, drawn twice with replacement.
Fill in the blanks
\text9 = 3 \times 3 = 121, \qquad \text___ = 11 \times 11 = ___
Why: Nine of the 121 outcomes are red then red. With replacement both stages have all eleven balls available, which is why both factors are 11 and both red factors are 3.
Worked example
Example 3.24c. Two paths, and they cannot both happen.
\[ P(RB \text{ OR } BR) \]
Identify the qualifying paths
Why: Red then blue, and blue then red.
Read their counts
Why: Twenty-four each.
\[ 24\text{ and } 24 \]
Add them
Why: The paths are mutually exclusive.
\[ 48 \]
Divide by the total
Why: Forty-eight of 121.
\[ \frac{48}{121} \]
Figure (svg): The solution to Worked example an OR from a tree shown as a ladder of expressions, one row per legal move
\[ P(RB \text{ OR } BR) = \frac{24}{121} + \frac{24}{121} = \frac{48}{121} \]
Verify: confirm why no subtraction was needed
Why: Section 3.3's addition rule subtracts the intersection, and here the intersection is empty: a pair of draws cannot be both red-then-blue and blue-then-red. Distinct paths through a tree are always mutually exclusive, so unions of paths are always plain sums. That is one of the tree's quiet advantages — it arranges the outcomes so that the correction term is guaranteed to be zero.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 190
Trap
\[ \text{one red and one blue: that is one outcome} \]
Count a single leaf for the mixed result
Why: The two orders give the same pair of colours, so they look like one case.
\[ P = \frac{24}{121} \quad \text{(half the true answer)} \]
The draws are made one at a time, so red-then-blue and blue-then-red are different sequences and the tree gives them different paths.
\[ P(\text{one of each}) = \frac{24 + 24}{121} = \frac{48}{121} \]
Count every path that satisfies the description
Why: Order matters in a sequential experiment, so a description ignoring order usually covers several paths.
This is the same error section 3.1 met with two coins, where collapsing HT and TH into 'one head' gave a sample space of three unequal outcomes. A tree prevents it structurally: each path is drawn separately, so a description covering two of them visibly requires two leaves to be added.
Sorting
Two draws with replacement, paths RR, RB, BR and BB.
Sort into buckets
Sort each description by how many paths it covers.
Item (d) is the case where the complement is quicker: at least one red is everything except BB, so the answer is 1 minus 64 over 121, which is 57 over 121. Adding the three qualifying leaves gives 9 plus 24 plus 24, the same 57.
Two truths and a lie
All three concern frequency trees.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Red-then-blue and blue-then-red are different paths with different first branches, even though both give one ball of each colour. The tree is a diagram of a sequence, so order is built into its structure.
Prediction
Commit before reasoning.
Predict first
Two balls are drawn from eleven, with replacement. Why is the sample space 121 rather than something smaller?
Correct: Each first-draw possibility pairs with all eleven second-draw possibilities.
Why: With replacement the urn is restored, so every one of the eleven first outcomes can be followed by any of eleven second outcomes, giving 11 times 11. The balls being distinguishable is what makes those 121 outcomes equally likely, which the book takes trouble to spell out by listing R1R1 through R3R3.
Section
Section 2
Concept
The branches may be labelled with probabilities instead of frequencies. The numbers at the ends of the branches are then calculated by multiplying the numbers on the two corresponding branches, so a path's probability is the product of the probabilities along it.
path probability — The probability of a complete path through the tree, found by multiplying the branch probabilities along it. Because the second branch carries a conditional probability, this is exactly section 3.3's general multiplication rule.
\[ P(R \text{ then } B) = P(R) \cdot P(B\mid R) = \frac{3}{11} \cdot \frac{8}{10} = \frac{24}{110} \]
The second branch is the important one. It carries the probability of the second draw GIVEN what the first draw was, which is why the tree handles dependent draws without any extra machinery. The book labels the general tree explicitly in this way, noting that P(R given B) on the diagram means the probability of red on the second draw given blue on the first. Multiplying along a path is therefore the general multiplication rule, not the independent special case.
Figure (svg): A diagram showing that the probability of a path through a tree is the product of the branch probabilities along it
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 190-192 — the probability tree, and the conditional labels
Picture it
The multiplication rule laid out along a branch.
Figure (svg): A diagram showing that the probability of a path through a tree is the product of the branch probabilities along it
Drawing the rule this way makes the conditional impossible to forget, which is its main pedagogical value. Section 3.3's commonest error was multiplying two unconditional probabilities; on a tree the second branch is physically labelled with a conditional, so the correct quantity is the one in front of you.
Worked example
Example 3.25. The same urn, drawn without replacement.
\[ 3 \text{ red}, 8 \text{ blue}; \text{ draw two without replacement} \]
First-stage branches
Why: Three of eleven, eight of eleven.
\[ \frac{3}{11}\text{ and } \frac{8}{11} \]
Second stage after a red
Why: Two reds left among ten balls.
\[ \frac{2}{10}\text{ and } \frac{8}{10} \]
Second stage after a blue
Why: Three reds still, among ten.
\[ \frac{3}{10}\text{ and } \frac{7}{10} \]
Multiply along each path
Why: Four products.
\[ \frac{6}{110}, \frac{24}{110}, \frac{24}{110}, \frac{56}{110} \]
Figure (svg): The solution to Worked example the tree without replacement shown as a ladder of expressions, one row per legal move
\[ P(RR) = \frac{3}{11}\cdot\frac{2}{10} = \frac{6}{110} \]
Verify: confirm the four paths total one
Why: Six plus 24 plus 24 plus 56 is 110, over a denominator of 110, so the four path probabilities sum to exactly one. They must, because the four paths are mutually exclusive and between them cover every possible pair of draws. That total is the probability tree's version of the frequency tree's leaf count, and it is the check to run before reading any answer off the diagram.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 190-191
Faded example
Without replacement from three red and eight blue.
Fill in the blanks
P(B \text3 R) = \frac24___ \cdot \frac___}___ = \frac___}___
Why: Taking a blue leaves all three reds among ten balls, so the second branch is 3/10 and the path probability is 24 over 110. Note this equals P(R then B), which was 3/11 times 8/10 — the same factors in the other order.
Worked example
The same urn, and the one place the two diagrams differ.
\[ \text{with replacement against without} \]
Compare the first stage
Why: Three of eleven and eight of eleven, in both.
Compare the second stage after red
Why: Three of eleven, against two of ten.
Say why
Why: Without replacement a red has been removed.
Note the consequence for P(RR)
Why: 9/121 against 6/110.
\[ 0.0744\text{ against } 0.0545 \]
Figure (svg): The solution to Worked example comparing the two trees shown as a ladder of expressions, one row per legal move
\[ \frac{9}{121} \approx 0.074 \quad \text{against} \quad \frac{6}{110} \approx 0.055 \]
Verify: confirm the direction of the change for each path
Why: Two reds becomes less likely without replacement, since a red has been removed; two blues also becomes less likely, for the same reason. The mixed paths become MORE likely — 24 over 110 is about 0.218 against 24 over 121, about 0.198 — because removing one colour makes the other more available. So the effect is not uniform, and it is worth checking each path rather than assuming the whole tree shifts one way.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 190-191
Trap
\[ \text{second-stage branches: } \frac{3}{11} \text{ and } \frac{8}{11} \]
Copy the first stage's numbers onto the second
Why: The two stages look symmetric, so the same labels seem right.
\[ \text{that is the WITH-replacement tree} \quad \text{(the urn has not been restored)} \]
Ten balls remain after the first draw, and how many are red depends on what the first ball was.
\[ \text{after R: } \frac{2}{10}, \frac{8}{10}; \qquad \text{after B: } \frac{3}{10}, \frac{7}{10} \]
Label each second-stage branch for the state the first draw left behind
Why: The two subtrees have DIFFERENT labels, which is the visual signature of dependence.
A quick diagnostic: on a with-replacement tree the two second-stage subtrees are identical, and on a without-replacement tree they differ. If you have drawn a without-replacement tree whose subtrees match, the second stage was not updated. Every denominator at the second stage should also have dropped by one.
Discrimination
Look at the second-stage branches.
Sort into buckets
Sort each description of a two-stage tree.
Two truths and a lie
All three concern probability trees.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false, and it is the point of the second branch being conditional. Multiplying along a path is the GENERAL multiplication rule, which needs no independence — that is precisely why the without-replacement tree works. Independence only makes the two subtrees identical.
Prediction
Commit before reasoning.
Predict first
You have drawn a probability tree for a two-stage experiment. What should the path probabilities sum to?
Correct: Exactly one.
Why: The paths are mutually exclusive and between them cover every possible outcome, so their probabilities partition the total probability of one. This holds for any tree with any number of stages and any dependence structure, which makes it the universal check — the without-replacement tree's 6, 24, 24 and 56 over 110 sum to one just as the with-replacement tree's leaves sum to 121 out of 121.
Section
Section 3
Concept
A conditional probability on a tree is read by restricting attention to the paths consistent with the condition. Conditioning on the first draw keeps one subtree; conditioning on the second draw requires collecting the qualifying paths and dividing by their total.
conditioning on a tree — Restricting to the paths consistent with the condition. If the condition concerns the first stage, the answer is a single second-stage branch. If it concerns the second stage, the qualifying path probabilities must be summed and divided by the total of all paths meeting the condition.
\[ P(R \text{ on 2nd} \mid B \text{ on 1st}) = \frac{3}{10} \quad \text{: one branch} \]
The two directions are asymmetric and that asymmetry is worth noticing. Conditioning on the FIRST stage is free — the answer is already written on a branch, because the tree was built in that order. Conditioning on the SECOND stage is work: the book's Example 3.24e collects the 24 BR and 64 BB outcomes to get a reduced sample space of 88, and reads 24 of those 88. A tree is built for one direction and reluctant in the other.
Figure (svg): A tree diagram shown twice, the second time with the red first-draw subtree faded out, leaving only the blue subtree from which the conditional is read
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 190-191 — conditional probabilities read from the tree
Picture it
Conditioning on blue first fades the entire red subtree.
Figure (svg): A tree diagram shown twice, the second time with the red first-draw subtree faded out, leaving only the blue subtree from which the conditional is read
The answer of three tenths was already printed on a branch before any conditioning was done, because the tree is organised by first draw. That is the sense in which a tree is the natural display for a sequential problem: the conditionals you are most likely to want are its labels.
Worked example
Example 3.25c. The answer is a branch label.
\[ P(R \text{ on 2nd} \mid B \text{ on 1st}), \text{ without replacement} \]
Identify the condition
Why: Blue on the first draw.
Discard the other subtree
Why: The red first-draw paths are impossible now.
Find the branch for red on the second
Why: Within the surviving subtree.
\[ \frac{3}{10} \]
Read it off
Why: No further arithmetic.
\[ \frac{3}{10} \]
Figure (svg): The solution to Worked example conditioning on the first draw shown as a ladder of expressions, one row per legal move
\[ P(R \text{ on 2nd}\mid B \text{ on 1st}) = \frac{3}{10} \]
Verify: confirm with the formula as a cross-check
Why: The formula gives P(BR) divided by P(B on first), which is 24 over 110 divided by 8 over 11. That is 24 over 110 times 11 over 8, which simplifies to 3 over 10 — the same answer. The tree got there without the division because the branch was already conditional, which is exactly the labour the diagram saves.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 191
Faded example
With replacement, the leaves are RR 9, RB 24, BR 24, BB 64.
Fill in the blanks
P(R \text64 \mid B \text88) = \frac______}} = \frac______}
Why: The condition keeps only the two blue-first leaves, totalling 88 outcomes, and 24 of them have red second. That simplifies to three elevenths, the unconditional probability, confirming independence under replacement.
Worked example
Example 3.24e, the harder direction, on the with-replacement tree.
\[ P(R \text{ on 2nd} \mid B \text{ on 1st}) \text{ from the frequency tree} \]
Collect the paths meeting the condition
Why: Blue first: the BR and BB leaves.
\[ 24\text{ and } 64 \]
Total them for the reduced sample space
Why: Twenty-four plus 64.
\[ 88\text{ outcomes} \]
Count those also meeting the event
Why: The BR leaf.
\[ 24 \]
Divide
Why: Twenty-four of 88.
\[ \frac{24}{88} = \frac{3}{11} \]
Figure (svg): The solution to Worked example conditioning on the second draw shown as a ladder of expressions, one row per legal move
\[ P = \frac{24}{24 + 64} = \frac{24}{88} = \frac{3}{11} \]
Verify: confirm the answer equals the unconditional probability
Why: Three elevenths is exactly P(red) on any single draw, so conditioning on a blue first draw changed nothing. That is what independence looks like on a tree, and it is expected here because this is the WITH-replacement diagram. On the without-replacement tree the same conditional came out at three tenths rather than three elevenths, and that difference is the dependence.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 190
Error analysis
Without replacement from three red and eight blue, a student wants P(R on 2nd given B on 1st). The four path probabilities are 6/110, 24/110, 24/110 and 56/110.
Annotate
On: \( \begin{aligned} &(1)\; \tfrac{24}{110} \\ &(2)\; \tfrac{3}{11} \\ &(3)\; \tfrac{24}{110} \div \tfrac{80}{110} \\ &(4)\; \tfrac{24}{110} \div \tfrac{80}{110} = \tfrac{24}{80} = \tfrac{3}{10} \end{aligned} \)
Error (1) is the joint-for-conditional confusion of section 3.1 in a new setting, and error (2) is the with-replacement answer used on a without-replacement tree. Both are avoided by reading the branch directly, which is what the tree was drawn for.
Discrimination
A tree built by first draw, then second draw.
Sort into buckets
Sort each conditional by how much work it takes to read.
Two truths and a lie
All three concern conditionals on a tree.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The tree is organised by the order of the stages, so conditioning on the second stage means collecting every path with that second outcome and dividing by their total — the book does exactly this work in Example 3.24e, reaching a reduced sample space of 88.
Prediction
Commit before reasoning.
Predict first
On the with-replacement tree, P(red on 2nd given blue on 1st) comes out at 3/11, the same as P(red). What does that show?
Correct: That the two draws are independent.
Why: Section 3.2's first condition for independence is that the conditional equals the unconditional probability, and that is exactly what has happened. It is expected here because the urn was restored, and the without-replacement tree gives three tenths instead — the difference between the two answers is the dependence that replacement removes.
Section
Section 4
Concept
A Venn diagram represents events as regions within the sample space, so that the overlap between two events is visible directly. Two events divide the sample space into four regions: in A only, in both, in B only, and in neither.
the four regions — Two events split the sample space into A only, A and B, B only, and neither. Once a count is placed in each region, every probability of the chapter is a sum of regions divided by the grand total.
\[ P(A \text{ OR } B) = \frac{\text{A only} + \text{both} + \text{B only}}{n} \]
The four regions are mutually exclusive and between them exhaustive, which is what makes the display work: every member of the sample space falls in exactly one of them, so counts simply add. That also explains the addition rule pictorially — adding the A circle and the B circle covers the overlap twice, and the correction is the region counted a second time.
Figure (svg): A Venn diagram of two overlapping events with counts in each of the four regions, used to read off unions, intersections and complements
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 193-195 — Venn diagrams for two events
Picture it
Fifty members, split into A only, both, B only and neither.
Figure (svg): A Venn diagram of two overlapping events with counts in each of the four regions, used to read off unions, intersections and complements
The working method is always to fill the four regions before answering anything, and to fill the OVERLAP first — because the count for A includes the overlap, so A only is A minus the overlap. Once the four numbers are in place, every question in the chapter is answered by adding some of them and dividing by fifty.
Worked example
The order matters: the overlap first, then the crescents.
\[ n = 50, \quad |A| = 25, \quad |B| = 19, \quad |A \text{ AND } B| = 7 \]
Place the overlap
Why: Given directly.
\[ \text{both } = 7 \]
Find A only
Why: A's total minus the overlap.
\[ 25 - 7 = 18 \]
Find B only
Why: B's total minus the overlap.
\[ 19 - 7 = 12 \]
Find neither
Why: The grand total minus the other three.
\[ 50 - 18 - 7 - 12 = 13 \]
Figure (svg): The solution to Worked example filling the regions shown as a ladder of expressions, one row per legal move
\[ 18 + 7 + 12 + 13 = 50 \]
Verify: confirm the four regions total the sample space
Why: They sum to exactly 50, which they must, since every member falls in one and only one region. That total is the Venn diagram's own check, exactly like the margins of a contingency table and the leaves of a tree — each of the three displays partitions the sample space, and each provides a total to check against.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 193-194
Faded example
Fifty members, with 25 in A, 19 in B and 7 in both.
Fill in the blanks
\text18 = 25 - 7 = 7, \qquad \text___ = 19 - ___ = 12
Why: Each crescent is its circle's total minus the shared part. Filling the overlap first and subtracting outward is the reliable order, because it makes double counting structurally impossible.
Worked example
Once filled, every answer is a sum of regions.
\[ \text{regions: A only } 18, \text{ both } 7, \text{ B only } 12, \text{ neither } 13 \]
P(A)
Why: A only plus the overlap.
\[ \frac{18 + 7}{50} = 0.50 \]
P(A AND B)
Why: The overlap alone.
\[ \frac{7}{50} = 0.14 \]
P(A OR B)
Why: All three shaded regions.
\[ \frac{18 + 7 + 12}{50} = 0.74 \]
P(neither)
Why: The region outside both circles.
\[ \frac{13}{50} = 0.26 \]
Figure (svg): The solution to Worked example reading probabilities off the regions shown as a ladder of expressions, one row per legal move
\[ P(A \text{ OR } B) = 0.74 = 1 - 0.26 = 1 - P(\text{neither}) \]
Verify: confirm the union and the outside are complements
Why: The union is 0.74 and the outside region is 0.26, and they sum to one — as they must, since 'at least one of A and B' and 'neither' are complementary events. That gives a second route to the union and a check on the first, and it is often the faster route: on a Venn diagram the outside region is a single number while the union is three added together.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 194-195
Trap
\[ |A| = 25, \text{ so write } 25 \text{ in the left crescent} \]
Write each event's total in its own crescent
Why: The count of 25 is A's number, so it goes in A's part of the picture.
\[ 25 + 7 + 12 = 44 \ne 50 \quad \text{(and A now has 32)} \]
The 25 already includes the seven members who are also in B, so writing 25 outside the overlap counts them twice.
\[ \text{A only} = 25 - 7 = 18 \]
Fill the overlap first, then subtract it from each event's total
Why: The circles are not the regions; a circle is two regions together.
This is the commonest error with Venn diagrams and it always announces itself: the four regions will not total the sample size. Filling the overlap first makes it impossible, because the two crescents are then computed by subtraction rather than written down. The same logic applies to a three-event diagram, where the centre region is filled first and everything else works outward.
Sorting
The four regions are A only, both, B only, and neither.
Sort into buckets
Sort each event by how many regions it covers.
Item (e), A but not B, is the crescent alone — and it is worth noticing that it is a single region while A itself is two. That distinction is exactly what the trap in this idea is about.
Two truths and a lie
All three concern Venn diagrams.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false as usually intended. A circle is made of two regions, and the numbers written on a filled diagram are region counts, not circle totals. The circle's total is its crescent plus the overlap, which is why the crescent is found by subtraction.
Prediction
Commit before reasoning.
Predict first
What goes wrong if you fill the crescents before the overlap?
Correct: You do not yet know how much belongs outside the overlap.
Why: Each circle's total is a crescent plus a shared part, so the crescent cannot be determined until the shared part is known. Filling the overlap first turns both crescents into subtractions and makes the diagram consistent by construction. Working inward-out is the general rule and it extends to three-event diagrams, where the central region is filled first.
Section
Section 5
Concept
The chapter now offers three displays. A tree suits a sequence of stages, a Venn diagram suits overlapping events on a single trial, and a contingency table suits two variables measured on one sample. The right choice is whichever makes the wanted quantity a single number to read.
choosing a display — A tree for sequential experiments, where the conditionals run forward through the stages. A Venn diagram for two or three overlapping events on one trial. A contingency table for two categorical variables, especially when either has more than two categories.
\[ \text{tree: stages} \qquad \text{Venn: overlaps} \qquad \text{table: two variables} \]
The three overlap considerably and a problem can often be drawn more than one way. What distinguishes them is which quantity comes out for free. A tree hands you path probabilities and forward conditionals; a Venn diagram hands you unions and complements; a table hands you conditionals in both directions. Choosing badly does not make a problem unsolvable, only longer.
Figure (svg): Two columns comparing the tree diagram, which suits a sequence of stages, with the Venn diagram, which suits overlapping events
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 189-195 — the two displays of this section, alongside section 3.4's table
Picture it
What each one is built to make easy.
Figure (svg): Two columns comparing the tree diagram, which suits a sequence of stages, with the Venn diagram, which suits overlapping events
The last row is the practical limit. A tree with three stages of three outcomes has 27 paths and stops being readable; a Venn diagram with four events cannot be drawn with circles at all. Both are tools for small problems, which is exactly the case where a picture helps most — the book's reason for the section is that graphing helps when a problem is complex, and these stop helping when it becomes complex in the wrong way.
Worked example
Two draws from an urn without replacement.
\[ \text{Find P(one of each colour) for two draws from } 3 \text{ red and } 8 \text{ blue} \]
Note the structure
Why: Two stages, with the second depending on the first.
Choose the display
Why: A tree carries conditionals on its second branches.
Read the qualifying paths
Why: Red then blue, and blue then red.
\[ \frac{24}{110}\text{ each} \]
Add them
Why: The paths are mutually exclusive.
\[ \frac{48}{110} \]
Figure (svg): The solution to Worked example choosing for a sequential problem shown as a ladder of expressions, one row per legal move
\[ P = \frac{24}{110} + \frac{24}{110} = \frac{48}{110} \]
Verify: confirm a Venn diagram would have been awkward here
Why: A Venn diagram has no natural way to express 'first' and 'second', so the two orderings would have to be handled outside the picture — losing the display's whole advantage. The tree makes the ordering structural, which is why the two paths appear separately and are simply added. Conversely, a question about two properties of a single ball would be awkward on a tree and natural on a Venn diagram.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, p. 191
Sorting
Ask whether the problem has stages, overlaps, or two variables with several categories.
Sort into buckets
Sort each problem.
Items (b) and (c) are structurally the same problem at different scales, which is why a two-by-two table and a two-event Venn diagram are interchangeable. Item (e) is where the table wins outright: three terrain categories cannot be drawn as overlapping circles.
Worked example
Two properties of one randomly chosen person.
\[ \text{Of } 50 \text{ people}, 25 \text{ in A}, 19 \text{ in B}, 7 \text{ in both. Find P(neither).} \]
Note the structure
Why: One trial, two properties that can co-occur.
Choose the display
Why: A Venn diagram shows the four regions at once.
Fill the regions
Why: Overlap first, then the crescents, then outside.
\[ 18, 7, 12, 13 \]
Read the answer
Why: The region outside both circles.
\[ \frac{13}{50} = 0.26 \]
Figure (svg): The solution to Worked example choosing for an overlap problem shown as a ladder of expressions, one row per legal move
\[ P(\text{neither}) = \frac{13}{50} = 0.26 \]
Verify: confirm with the complement of the union
Why: The union is 18 plus 7 plus 12 over 50, which is 0.74, and one minus 0.74 is 0.26. The two routes agree. Note that a tree would have handled this badly — there is no sequence here, only one person with two properties — while a two-by-two contingency table would have worked perfectly well, with the four regions as its four cells. Venn and table are close substitutes; the tree is the odd one out.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 193-195
Trap
\[ \text{two draws: let A = red, B = red, and draw a Venn diagram} \]
Represent the two draws as two overlapping events
Why: There are two events, so a two-event diagram seems appropriate.
\[ \text{the diagram cannot express WHICH draw was which} \]
Red-then-blue and blue-then-red would fall in the same region, and the ordering that the problem is about has been erased.
\[ \text{use a tree: first stage, then second stage} \]
Ask whether the problem has stages or properties
Why: Stages need a tree; properties of one trial need a Venn diagram or a table.
The diagnostic question is whether the word 'then' belongs in the description. Draw a ball, THEN draw another is sequential and wants a tree. Choose a person who may be tall AND may be left-handed has no 'then' and wants a Venn diagram or a table. Problems that genuinely have both structures — several stages, each with several properties — usually need a tree whose leaves are then tabulated.
Matching
Each display hands you one quantity without work.
Match the pairs
Why: The fourth row is the property all three share and it is worth relying on: a tree's leaves sum to one, a Venn diagram's regions sum to the sample size, and a table's margins sum to the grand total. Each display partitions the sample space, and each therefore supplies its own arithmetic check.
Two truths and a lie
All three concern choosing a display.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A Venn diagram has no way to express order, so a sequential problem loses exactly what it is about, and circles cannot properly represent four or more events. Each display has a shape of problem it suits and shapes it cannot handle.
Explain it
A classmate has drawn a Venn diagram for two draws from an urn and cannot make it work.
Discussion prompt
In three sentences or fewer, diagnose the problem and say what to draw instead.
Hint: Ask where the ordering lives on their diagram.
Answer:
Ask them where red-then-blue sits on their picture, and then where blue-then-red sits — they will find both fall in the same region, so the diagram has erased the ordering the problem is about.
A Venn diagram shows which properties a single outcome has, and it has no way to say which came first.
Draw a tree instead: the first set of branches is the first draw, the second set is the second, and the two orderings become two separate paths whose probabilities can be added.
Comparison
Fill the blanks. The last column is the check each one supplies.
Comparison matrix
| Display | Suits | Its own total |
|---|---|---|
| Tree diagram | a sequence of stages | the path probabilities sum to one |
| Venn diagram | overlapping events on a single trial | the four regions sum to the sample size |
| Contingency table | two variables measured on one sample | both margins sum to the grand total |
| What each makes free | tree: forward conditionals; Venn: unions; table: both conditionals | all three: a partition, and so a check |
Every one of the three partitions the sample space, which is what makes their counts add and what gives each a total to check against. Choosing between them is choosing which quantity you want handed to you rather than computed.
Pattern
Six steps. Step three is where without-replacement problems go wrong.
For a Venn diagram the order is the mirror image: fill the overlap first, subtract to get each crescent, then subtract everything from the sample size to get the outside region, and check that the four regions total n.
OpenStax Introductory Business Statistics 2e, §3.5 Venn Diagrams §3.5 Venn Diagrams
Check
Multiply along the path.
Check your understanding
An urn holds 3 red and 8 blue. Two are drawn without replacement. What is P(red then blue)?
Answer: A
Why: The first branch is 3 of 11, and after a red is removed ten balls remain of which eight are blue, so the second branch is 8 of 10.
Check
Read a conditional.
Check your understanding
On the without-replacement tree, what is P(red on the second draw given blue on the first)?
Answer: A
Why: Conditioning on blue first keeps only that subtree, and within it three reds remain among ten balls, so the branch reads 3/10.
Check
Fill the Venn regions.
Check your understanding
Of 50 people, 25 are in A, 19 in B and 7 in both. How many are in A only?
Answer: A
Why: The 25 in A includes the 7 who are also in B, so A only is 25 minus 7, which is 18.
Real world
A factory has two suppliers. Supplier A provides 70 percent of components with a 2 percent defect rate; supplier B provides the remaining 30 percent with a 5 percent defect rate. A component is pulled at random from the line and found to be defective.
Discussion prompt
Draw the situation, find the probability that it came from supplier A, and say why the answer is not 70 percent.
Hint: The tree runs supplier then defect; the question runs the other way.
Answer:
Draw the tree with supplier first. The four paths are A-defective at 0.70 times 0.02, or 0.014; A-good at 0.686; B-defective at 0.30 times 0.05, or 0.015; and B-good at 0.285. The four sum to one, which checks the tree.
The question runs backwards through the stages. It conditions on the second stage, so the qualifying paths must be collected: the defectives total 0.014 plus 0.015, which is 0.029. Of that, A's share is 0.014.
So P(A given defective) is 0.014 over 0.029, about 0.483 — under half, even though A supplies 70 percent of the components. B's higher defect rate more than compensates for its smaller share, so a defective component is very nearly as likely to have come from the smaller supplier.
\[ P(A\mid\text{defective}) = \frac{0.014}{0.014 + 0.015} \approx 0.483 \]
This is the backwards-conditional case the third idea flagged as the tree's awkward direction, and it is the most useful one in practice — quality control, medical testing and fault diagnosis all ask which source produced an observed effect. The general form of this calculation is called Bayes' theorem, and the tree is where it becomes obvious: collect the paths consistent with what you observed, and divide.
Commit first
Answer, then rate your confidence honestly.
Predict first
What does the second set of branches on a probability tree carry?
Correct: Conditional probabilities.
\[ P(\text{path}) = P(\text{first}) \cdot P(\text{second} \mid \text{first}) \]
Why: Each second-stage branch is labelled with the probability of that outcome GIVEN what happened at the first stage, which is why multiplying along a path is the general multiplication rule and works for dependent draws. The book labels its general tree exactly this way. With replacement those conditionals happen to equal the unconditional probabilities, which is why the two subtrees look identical there and only there.
Explain it
They have drawn a without-replacement tree and copied the first stage's probabilities onto the second.
Discussion prompt
In three sentences or fewer, show them the error using the shape of their own drawing.
Hint: Ask them to compare the two subtrees.
Answer:
Point out that their two subtrees are identical, and ask what that would mean: that the second draw is the same whether a red or a blue was removed.
Without replacement ten balls remain rather than eleven, and how many are red depends on what went — so after a red the branch is 2 of 10 and after a blue it is 3 of 10.
Identical subtrees are the signature of independence, so a without-replacement tree whose subtrees match has quietly become a with-replacement tree.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the second stage, update both the numerator and the denominator for what the first draw removed, and check the two subtrees differ. For a backwards conditional, collect every path meeting the condition and divide. For a Venn diagram, fill the overlap first and subtract outward. For choosing a display, ask whether the word 'then' belongs in the problem. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top left, draw the tree for two draws from an urn of three red and eight blue WITH replacement, labelling every branch with a frequency and every leaf with an outcome count, and check that the four leaves total 121. Beside it, draw the tree for the SAME urn WITHOUT replacement, labelling every branch with a probability and every leaf with a path probability, and check that the four paths total one. Circle the one place the two trees differ and write one sentence saying what that difference is. In the middle, use the second tree to answer four questions, showing your working: P(RR), P(one of each colour), P(red on the second given blue on the first), and P(blue on the first given red on the second) — noting which of the four was free and which took work. At the bottom left, draw a two-event Venn diagram for 50 people with 25 in A, 19 in B and 7 in both: fill all four regions in the right order and check they total 50, then read off P(A), P(A AND B), P(A OR B) and P(neither). At the bottom right, write a three-row table naming each display, the shape of problem it suits, and the total it supplies as a check.
Check the two trees against each other: the first-stage branches should be identical and every second-stage branch should differ. If any second-stage branch matches between the two diagrams, the without-replacement tree was not updated.
Recap
Two displays, and the judgement about when each is worth drawing.
| If you see | Then |
|---|---|
| A sequence of draws or stages | Tree diagram |
| Two properties of a single trial | Venn diagram or a two-by-two table |
| Two variables with several categories | Contingency table |
| Drawing without replacement | Update both parts of every second-stage branch |
| Two identical subtrees | The stages are independent |
| A condition on the first stage | The answer is a branch label |
| A condition on the second stage | Collect the qualifying paths and divide |
Chapter 3 is finished: the vocabulary, the two relationships events can have, the two rules, and three displays for applying them. Chapter 4 turns to random variables — a number attached to each outcome — and asks not what a single event's probability is but how probability is distributed across all the values a quantity can take.
OpenStax Introductory Statistics 2e, §3.5 Tree and Venn Diagrams §3.5, pp. 189-195 — everything on these slides traces back here
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