3.5 Tree and Venn Diagrams
SLO 3
Describe and apply probability concepts and distributions.
The rules from the earlier sections get a picture here. A tree lays out every path of a staged experiment, so you multiply along one path and add across several; a Venn diagram turns AND, OR, and 'neither' into regions you can point at and read a probability off.
Learning Objectives
By the end of this section, you will be able to:
- build a tree diagram for a two-stage experiment and label its branches with either frequencies or probabilities;
- read a probability off a tree by multiplying along one path and adding across paths;
- explain how drawing with replacement and drawing without replacement change the second set of branches;
- draw a Venn diagram for two events and shade the regions that stand for AND, OR, and neither;
- pull \(P(A \text{ AND } B)\), \(P(A \text{ OR } B)\), and \(P(A \mid B)\) off a completed diagram.
Sections 3.3 and 3.4 gave you formulas. This section gives you pictures. When a probability problem gets tangled enough that you cannot tell which rule to reach for, the fastest way out is usually to draw the situation instead of writing more symbols.
Two pictures do almost all of the work. A tree diagram is the right picture when something happens in stages — draw a card, then draw another. A Venn diagram is the right picture when everything happens at once and you only need to know how two groups overlap. Both of them turn AND, OR, and "given" from words you have to parse into places on a page you can point at.
3.5.1 Tree Diagrams
A tree diagram is a graph that shows every possible outcome of a multi-stage experiment. Each stage of the experiment gets its own set of branches, and each branch is labeled with either the frequency (how many ways that step can happen) 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.
Two facts make a tree usable, and they are the only two you need to memorize:
- 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.
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.
Each fork is a moment where the experiment could break one way or the other, and each tip is one complete story from start to finish. Multiplying along a path is asking "how likely is this exact story?" Adding paths is asking "how likely is any story that ends this way?"
Definition 3.5.1 — A tree diagram: multiply along one path, add across several.
Definition 3.5.2 — With replacement: the ball goes back, so draw 2 faces the same urn.
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.
The difference between those two definitions is the whole reason trees are worth drawing. In the with-replacement case every second-stage fork looks the same, so you could get by with a formula. In the without-replacement case each second-stage fork carries its own numbers, and keeping them straight in your head is exactly the kind of bookkeeping people get wrong. The tree does the bookkeeping for you: you write the shrinking counts on the branches once, and then every question about the experiment is a matter of finding paths and multiplying.
Definition 3.5.3 — Without replacement: the ball stays out, so draw 2 faces a smaller urn.
In a standard deck there are 52 cards. Twelve of them are face cards (event \(F\)) and the other 40 are not face cards (event \(N\)). Draw two cards, one at a time, with replacement.
Using a tree diagram of frequencies, calculate \(P(FF)\).
Solution
Step 1 — build the four paths. Because the first card goes back in the deck, the second draw also has 12 face cards and 40 non-face cards available. Multiply the two branch counts to get the number of ways each path can happen:
| First card | Second card | Outcome | Number of ways |
|---|---|---|---|
| F (12) | F (12) | FF | 144 |
| F (12) | N (40) | FN | 480 |
| N (40) | F (12) | NF | 480 |
| N (40) | N (40) | NN | 1,600 |
Step 2 — find the size of the sample space. There are \(52(52) = 2{,}704\) outcomes, which is also what the four path counts add to: \(144 + 480 + 480 + 1{,}600 = 2{,}704\).
Step 3 — read off the FF path.
$$P(FF) = \frac{144}{2{,}704} = \frac{9}{169} \approx 0.0533$$Answer: \(P(FF) = \frac{9}{169} \approx 0.0533\).
An urn holds 11 balls: three are red (\(R\)) and eight are blue (\(B\)). Draw two balls, one at a time, with replacement — the first ball goes back in before the second is drawn.
The tree has two sets of branches. The first set is the first draw, the second set is the second draw, and because the ball was replaced both sets carry the same counts: 3 ways to get red, 8 ways to get blue. Multiplying the branch counts along each path gives the number of ways that path can happen.
| 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 |
The four path counts add to \(64 + 24 + 24 + 9 = 121\), which is the size of the sample space: there are \(11(11) = 121\) outcomes in all.
Each of those outcomes is distinct. If we label the red balls \(R_1, R_2, R_3\) and the blue balls \(B_1\) through \(B_8\), the nine \(RR\) outcomes can be listed one by one:
\(R_1R_1\); \(R_1R_2\); \(R_1R_3\); \(R_2R_1\); \(R_2R_2\); \(R_2R_3\); \(R_3R_1\); \(R_3R_2\); \(R_3R_3\)
The other three groups of outcomes are listed the same way.
Answer the following.
a) List the 24 \(BR\) outcomes: \(B_1R_1\), \(B_1R_2\), \(B_1R_3\), …
b) Find \(P(RR)\).
c) Find \(P(RB \text{ OR } BR)\).
d) Find \(P(R \text{ on 1st draw AND } B \text{ on 2nd draw})\).
e) Find \(P(R \text{ on 2nd draw} \mid B \text{ on 1st draw})\).
f) Find \(P(BB)\).
g) Find \(P(B \text{ on 2nd draw} \mid R \text{ on 1st draw})\).
Solution
Part a — list the paths that start blue and end red. Each of the eight blue balls can pair with each of the three red balls, which is \(8(3) = 24\) outcomes:
\(B_1R_1\); \(B_1R_2\); \(B_1R_3\); \(B_2R_1\); \(B_2R_2\); \(B_2R_3\); \(B_3R_1\); \(B_3R_2\); \(B_3R_3\); \(B_4R_1\); \(B_4R_2\); \(B_4R_3\); \(B_5R_1\); \(B_5R_2\); \(B_5R_3\); \(B_6R_1\); \(B_6R_2\); \(B_6R_3\); \(B_7R_1\); \(B_7R_2\); \(B_7R_3\); \(B_8R_1\); \(B_8R_2\); \(B_8R_3\)
Part b — multiply along the RR path.
$$P(RR) = \left(\frac{3}{11}\right)\left(\frac{3}{11}\right) = \frac{9}{121}$$Part c — two different paths satisfy "one of each color", so add them.
$$P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{11}\right) = \frac{48}{121}$$Part d — this names one specific path, red then blue.
$$P(R \text{ on 1st AND } B \text{ on 2nd}) = P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) = \frac{24}{121}$$Part e — a conditional question shrinks the sample space. We are told the first draw was blue, so only the two paths that start blue are still in play: \(BR\) with 24 outcomes and \(BB\) with 64. That leaves \(24 + 64 = 88\) possible outcomes, and 24 of them have red on the second draw.
$$P(R \text{ on 2nd} \mid B \text{ on 1st}) = \frac{24}{88} = \frac{3}{11}$$Part f — multiply along the BB path.
$$P(BB) = \left(\frac{8}{11}\right)\left(\frac{8}{11}\right) = \frac{64}{121}$$Part g — condition on a red first draw this time. The paths that start red are \(RR\) with 9 outcomes and \(RB\) with 24, so the reduced sample space has \(9 + 24 = 33\) outcomes. Twenty-four of those have blue on the second draw.
$$P(B \text{ on 2nd} \mid R \text{ on 1st}) = \frac{24}{33} = \frac{8}{11}$$Notice that parts e and g both came out equal to the plain probability of that color on a single draw. That is what replacement buys you: the two draws are independent, so knowing the first tells you nothing about the second.
Answer: a. the 24 listed \(BR\) outcomes; b. \(\frac{9}{121}\); c. \(\frac{48}{121}\); d. \(\frac{24}{121}\); e. \(\frac{3}{11}\); f. \(\frac{64}{121}\); g. \(\frac{8}{11}\).
Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?
Solution
Step 1 — count the pool. There are \(4 + 3 = 7\) balls. Because the first ball stays out, every second-stage denominator is 6.
Step 2 — find the two paths that give one of each color. Red then yellow, or yellow then red:
$$P(RY) = \left(\frac{4}{7}\right)\left(\frac{3}{6}\right) = \frac{12}{42}$$ $$P(YR) = \left(\frac{3}{7}\right)\left(\frac{4}{6}\right) = \frac{12}{42}$$Step 3 — add them.
$$P(\text{one of each}) = \frac{12}{42} + \frac{12}{42} = \frac{24}{42} = \frac{4}{7} \approx 0.571$$Answer: \(\frac{4}{7} \approx 0.571\).
Same urn: three red marbles and eight blue marbles. This time draw two marbles one at a time without replacement — you do not put the first marble back before selecting the second.
Now the second set of branches changes depending on which way the first draw went. If the first marble was red, only two reds are left among the ten remaining marbles. If the first marble was blue, all three reds are still there among the ten remaining. Labeling the branches with probabilities instead of frequencies:
| 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 | \(\frac{56}{110}\) |
The number at the end of each path is the product of the two branches on the way there — for instance \(\left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}\).
You drew a marble and did not put it back, so there are ten marbles in the urn when you reach in the second time. 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.
Use the tree to calculate the following.
a) \(P(RR)\)
b) Fill in the blanks: \(P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + (\underline{\phantom{xx}})(\underline{\phantom{xx}}) = \frac{48}{110}\)
c) \(P(R \text{ on 2nd} \mid B \text{ on 1st})\)
d) Fill in the blanks: \(P(R \text{ on 1st AND } B \text{ on 2nd}) = P(RB) = (\underline{\phantom{xx}})(\underline{\phantom{xx}}) = \frac{24}{110}\)
e) \(P(BB)\)
f) \(P(B \text{ on 2nd} \mid R \text{ on 1st})\)
Solution
Part a — multiply along the RR path, remembering the urn shrank.
$$P(RR) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}$$Part b — the missing factors are the BR branches.
$$P(RB \text{ OR } BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{24}{110} + \frac{24}{110} = \frac{48}{110}$$Part c — read the branch directly. A conditional probability on a tree is just a second-stage branch. Given a blue first draw, three of the ten remaining marbles are red:
$$P(R \text{ on 2nd} \mid B \text{ on 1st}) = \frac{3}{10}$$Part d — the missing factors are the two RB branches.
$$P(R \text{ on 1st AND } B \text{ on 2nd}) = P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{110}$$Part e — multiply along the BB path.
$$P(BB) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right) = \frac{56}{110}$$Part f — read the branch directly again. Given a red first draw, eight of the ten remaining marbles are blue:
$$P(B \text{ on 2nd} \mid R \text{ on 1st}) = \frac{8}{10} = \frac{4}{5}$$Compare part f here with part g of Example 3.5.1. 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, which is exactly what "dependent" means.
Answer: a. \(\frac{6}{110}\); b. \(\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)\); c. \(\frac{3}{10}\); d. \(\left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\); e. \(\frac{56}{110}\); f. \(\frac{8}{10}\).
If we are working in probabilities rather than counts, we can label any two-stage tree the same general way: first-stage probabilities on the first set of branches, conditional second-stage probabilities on the second set, and the product of the two written at each tip.
Make the tool compute the wrong urn on purpose
Open rāSHio, choose Calculators → Probability, switch the mode to Conditional, and enter \(P(A) = 0.2727\) and \(P(B) = 0.2727\) — both of them \(\frac{3}{11}\), the chance of drawing red — with Assume independence ticked. The panel returns \(P(A \text{ AND } B) = 0.07438\), which is \(\frac{9}{121}\), and it reports \(P(A \mid B) = 0.2727\), unchanged from \(P(A)\). That is Example 3.5.1's answer, not this one. Ticking Assume independence tells the tool to multiply two unchanged rates, which is precisely what putting the ball back does. The tree above gives \(\frac{6}{110} = 0.0545\) instead, and the gap between 0.0744 and 0.0545 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.
Figure 3.5.1 — rāSHio's Calculators → Probability panel in Conditional mode: with Assume independence ticked it returns the AND as the product of two unchanged rates, which is the with-replacement answer.
A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. Grant and his husband Peter come in, and they randomly select two kittens, without replacement, for adoption.
a) What is the probability that both kittens are tabby?
i. \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\) ii. \(\left(\frac{4}{9}\right)\left(\frac{4}{9}\right)\) iii. \(\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)\) iv. \(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)\)
b) What is the probability that one kitten of each coloring is selected?
i. \(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)\) ii. \(\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)\) iii. \(\left(\frac{4}{9}\right)\left(\frac{5}{9}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{9}\right)\) iv. \(\left(\frac{4}{9}\right)\left(\frac{5}{8}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\)
c) What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
d) What is the probability of choosing two kittens of the same color?
Solution
Set up the tree first. There are nine kittens: four tabby (\(T\)) and five black (\(K\)). Because the first kitten is not returned to the litter, every second-stage denominator is 8.
| 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 | \(\frac{20}{72}\) |
Part a — the TT path. After a tabby is taken, three tabbies remain out of eight kittens, so the answer is choice iii, \(\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)\). Choices i and ii both ignore the shrinking litter.
Part b — two paths give one of each color, so add them. The answer is choice iv:
$$P(\text{one of each}) = \left(\frac{4}{9}\right)\left(\frac{5}{8}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{8}\right) = \frac{20}{72} + \frac{20}{72} = \frac{40}{72}$$Part c — read the second-stage branch under "black first." A black kitten is gone, so eight kittens remain and all four tabbies are still there:
$$P(T \text{ on 2nd} \mid K \text{ on 1st}) = \frac{4}{8} = \frac{1}{2}$$Part d — add the two same-color paths.
$$P(\text{same color}) = \left(\frac{4}{9}\right)\left(\frac{3}{8}\right) + \left(\frac{5}{9}\right)\left(\frac{4}{8}\right) = \frac{12}{72} + \frac{20}{72} = \frac{32}{72} = \frac{4}{9}$$Answer: a. iii; b. iv; c. \(\frac{4}{8} = \frac{1}{2}\); d. \(\frac{32}{72} = \frac{4}{9}\).
3.5.2 Venn Diagrams
A Venn diagram is a picture of the outcomes of an experiment. It consists of a box that stands for the sample space \(S\), together with circles or ovals drawn inside the box. Each circle stands for one event, and an outcome is placed inside a circle exactly when it belongs to that event.
Once the picture is drawn, the three questions you keep asking about a pair of events all become places you can point to:
- \(A \text{ AND } B\) is the overlap where the two circles cross.
- \(A \text{ OR } B\) is everything inside either circle, overlap included but counted once.
- The region inside the box but outside both circles holds the outcomes in neither event.
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.
That last point is the one worth slowing down on. 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. When a problem tells you that 40% of students belong to a club and 50% work part time, the diagram has to leave room for the students who do neither, and the size of that leftover region is often the thing the question is really asking about.
Definition 3.5.4 — A Venn diagram: AND, OR, and neither are three regions you can point at.
Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event \(C = \{\text{green, blue, purple}\}\) and event \(P = \{\text{red, yellow, blue}\}\). Then \(C \text{ AND } P = \{\text{blue}\}\) and \(C \text{ OR } P = \{\text{green, blue, purple, red, yellow}\}\). Draw a Venn diagram representing this situation.
Solution
Step 1 — draw the box and label it \(S\). It holds all eight outcomes: black, white, red, orange, yellow, green, blue, purple.
Step 2 — draw two overlapping circles inside the box, one labeled \(C\) and one labeled \(P\).
Step 3 — place each outcome exactly once.
- In the overlap (both \(C\) and \(P\)): blue.
- In \(C\) only: green, purple.
- In \(P\) only: red, yellow.
- In the box but outside both circles: black, white, orange.
Step 4 — check the count. \(1 + 2 + 2 + 3 = 8\) outcomes, which is the whole sample space. Every outcome landed in exactly one region, so the diagram is complete.
Answer: a box labeled \(S\) containing two overlapping circles — blue in the overlap, green and purple in \(C\) alone, red and yellow in \(P\) alone, and black, white, and orange outside both.
Suppose an experiment has the outcomes 1, 2, 3, …, 12, where each outcome has an equal chance of occurring. Let event \(A = \{1, 2, 3, 4, 5, 6\}\) and event \(B = \{6, 7, 8, 9\}\). Draw the Venn diagram, and identify \(A \text{ AND } B\) and \(A \text{ OR } B\).
Solution
Step 1 — find the overlap. The only outcome that appears in both lists is 6:
$$A \text{ AND } B = \{6\}$$Step 2 — find the union. Everything in either list, written once:
$$A \text{ OR } B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$$Step 3 — place all twelve outcomes. Inside a box labeled \(S\), draw circle \(A\) and circle \(B\) overlapping.
- Overlap: 6
- \(A\) only: 1, 2, 3, 4, 5
- \(B\) only: 7, 8, 9
- Outside both circles: 10, 11, 12
Step 4 — sanity check with probabilities. Since all twelve outcomes are equally likely,
$$P(A \text{ OR } B) = \frac{9}{12} = 0.75 \qquad P(A \text{ AND } B) = \frac{1}{12} \approx 0.083$$and the addition rule agrees: \(\frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12}\).
Answer: \(A \text{ AND } B = \{6\}\); \(A \text{ OR } B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\); outcomes 10, 11, and 12 sit in the box outside both circles.
Roll a fair, six-sided die. Let \(A\) = a prime number of dots is rolled and \(B\) = an odd number of dots is rolled. Then \(A = \{2, 3, 5\}\) and \(B = \{1, 3, 5\}\), so \(A \text{ AND } B = \{3, 5\}\) and \(A \text{ OR } B = \{1, 2, 3, 5\}\). The sample space is \(S = \{1, 2, 3, 4, 5, 6\}\). Draw a Venn diagram representing this situation.
Solution
Step 1 — draw the box labeled \(S\) holding 1, 2, 3, 4, 5, 6, with two overlapping circles labeled \(A\) and \(B\) inside it.
Step 2 — place each face value once.
- Overlap (prime and odd): 3, 5
- \(A\) only (prime but not odd): 2
- \(B\) only (odd but not prime): 1
- Outside both circles: 4, 6
Step 3 — check the count. \(2 + 1 + 1 + 2 = 6\), the whole sample space.
Answer: 3 and 5 in the overlap, 2 in \(A\) alone, 1 in \(B\) alone, and 4 and 6 outside both circles.
Flip two fair coins. Let \(A\) = tails on the first coin, and let \(B\) = tails on the second coin. Draw the Venn diagram and identify \(A \text{ AND } B\), \(A \text{ OR } B\), and any outcome that belongs to neither event.
Solution
Step 1 — write the sample space. Reading each outcome as (first coin, second coin):
$$S = \{TT, TH, HT, HH\}$$Step 2 — list each event. Event \(A\) needs a tail in the first position, and event \(B\) needs a tail in the second position:
$$A = \{TT, TH\} \qquad B = \{TT, HT\}$$Step 3 — find the overlap and the union.
$$A \text{ AND } B = \{TT\} \qquad A \text{ OR } B = \{TT, TH, HT\}$$Step 4 — place all four outcomes. \(TT\) goes in the overlap, \(TH\) in \(A\) alone, \(HT\) in \(B\) alone, and \(HH\) sits inside the box but outside both circles — it is in neither \(A\) nor \(B\), because it has no tails at all.
Answer: \(A \text{ AND } B = \{TT\}\); \(A \text{ OR } B = \{TT, TH, HT\}\); \(HH\) belongs to neither event.
Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, and 5% work a second job and have a spouse who also works. Let \(J\) = works a second job and \(W\) = spouse also works. Draw a Venn diagram showing the relationships.
Solution
Step 1 — write down what you are given.
$$P(J) = 0.50 \qquad P(W) = 0.25 \qquad P(J \text{ AND } W) = 0.05$$Step 2 — peel the overlap out of each circle. The 5% in the overlap is already counted inside both the 50% and the 25%, so subtract it to get the "only" regions:
$$P(J \text{ only}) = 0.50 - 0.05 = 0.45 \qquad P(W \text{ only}) = 0.25 - 0.05 = 0.20$$Step 3 — find the leftover region. Everything inside the box has to add to 1:
$$P(\text{neither}) = 1 - (0.45 + 0.05 + 0.20) = 0.30$$Step 4 — label the diagram. A box labeled \(S\) with two overlapping circles: 0.45 in \(J\) alone, 0.05 in the overlap, 0.20 in \(W\) alone, and 0.30 in the box outside both circles.
Answer: the four regions carry 0.45, 0.05, 0.20, and 0.30, and they add to 1.
Forty percent of the students at a local college belong to a club, and 50% work part time. Five percent of the students work part time and belong to a club. Let \(C\) = student belongs to a club and \(PT\) = student works part time.
Draw a Venn diagram showing the relationships, then find the following for a randomly selected student.
a) The probability that the student belongs to a club.
b) The probability that the student works part time.
c) The probability that the student belongs to a club AND works part time.
d) The probability that the student belongs to a club GIVEN that the student works part time.
e) The probability that the student belongs to a club OR works part time.
Solution
Step 1 — build the diagram region by region. Start from the overlap and work outward, so nothing gets counted twice:
- Overlap (\(C\) and \(PT\)): 0.05
- \(C\) only: \(0.40 - 0.05 = 0.35\)
- \(PT\) only: \(0.50 - 0.05 = 0.45\)
- Neither: \(1 - (0.35 + 0.05 + 0.45) = 0.15\)
Part a — the whole \(C\) circle.
$$P(C) = 0.40$$Part b — the whole \(PT\) circle.
$$P(PT) = 0.50$$Part c — the overlap.
$$P(C \text{ AND } PT) = 0.05$$Part d — shrink the world to the \(PT\) circle. We are told the student works part time, so the \(PT\) circle is now the entire sample space, and we ask what fraction of it also lies in \(C\):
$$P(C \mid PT) = \frac{P(C \text{ AND } PT)}{P(PT)} = \frac{0.05}{0.50} = 0.1$$Part e — everything inside either circle. Add the two circles and subtract the overlap once, because adding them counted it twice:
$$P(C \text{ OR } PT) = P(C) + P(PT) - P(C \text{ AND } PT) = 0.40 + 0.50 - 0.05 = 0.85$$That matches the diagram directly: \(0.35 + 0.05 + 0.45 = 0.85\), which is also \(1 - 0.15\).
Answer: a. 0.40; b. 0.50; c. 0.05; d. 0.1; e. 0.85.
In a bookstore, the probability that a customer buys a novel is 0.6 and the probability that a customer buys a non-fiction book is 0.4. The probability that a customer buys both is 0.2.
a) Draw a Venn diagram representing the situation.
b) Find the probability that the customer buys either a novel or a non-fiction book.
c) In the Venn diagram, describe the overlapping area using a complete sentence.
d) Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.
Solution
Part a — build the four regions. Let \(N\) = buys a novel and \(F\) = buys a non-fiction book.
- Overlap: 0.2
- \(N\) only: \(0.6 - 0.2 = 0.4\)
- \(F\) only: \(0.4 - 0.2 = 0.2\)
- Neither: \(1 - (0.4 + 0.2 + 0.2) = 0.2\)
Part b — add the circles and remove the double count.
$$P(N \text{ OR } F) = 0.6 + 0.4 - 0.2 = 0.8$$Part c — say what the overlap means in words. The overlapping area represents the customers who buy both a novel and a non-fiction book.
Part d — where the new oval goes. Buying only compact disks means buying neither a novel nor a non-fiction book, so the new oval is drawn inside the box but not touching either of the existing circles. It sits in the 0.2 "neither" region.
Answer: a. four regions carrying 0.4, 0.2, 0.2, and 0.2; b. 0.8; c. customers who buy both a novel and a non-fiction book; d. a separate oval inside the box, disjoint from both circles.
A person with type O blood and a negative Rh factor (Rh-) can donate blood to a person with any blood type. Among African Americans, 4% have type O blood and a negative Rh factor, between 5% and 10% have the Rh- factor, and 51% have type O blood.
Take the average of 5% and 10% and use 7.5% as the percent with the Rh- factor. Let \(O\) = African American with type O blood and \(R\) = African American with the Rh- factor.
a) Find \(P(O)\).
b) Find \(P(R)\).
c) Find \(P(O \text{ AND } R)\).
d) Find \(P(O \text{ OR } R)\).
e) In the Venn diagram, describe the overlapping area using a complete sentence.
f) In the Venn diagram, describe the area inside the rectangle but outside both the circle and the oval, using a complete sentence.
Solution
Step 1 — draw the picture. The rectangle is the population of African Americans. The "O" circle holds those with type O blood; the "Rh-" oval holds those with the Rh- factor. They overlap because a person can have both.
Part a — read the circle.
$$P(O) = 0.51$$Part b — read the oval, using the 7.5% midpoint we were told to take.
$$P(R) = 0.075$$Part c — read the overlap.
$$P(O \text{ AND } R) = 0.04$$Part d — add the two regions and subtract the overlap once.
$$P(O \text{ OR } R) = P(O) + P(R) - P(O \text{ AND } R) = 0.51 + 0.075 - 0.04 = 0.545$$Part e — the overlap in words. The overlapping area represents the African Americans who have both type O blood and the Rh- factor — the universal donors this example opened with.
Part f — the leftover region in words. The area inside the rectangle but outside both shapes represents the African Americans who have neither type O blood nor the Rh- factor. Its probability is \(1 - 0.545 = 0.455\).
Answer: a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. those with both type O blood and the Rh- factor; f. those with neither type O blood nor the Rh- factor.
Problem Set 3.5
Problem 1. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result — meaning the test comes back positive for cancer when the man does not have it — is 0.51. Let \(C\) = a man develops cancer in his lifetime and \(F\) = a man has at least one false positive. Construct a tree diagram of the situation.
Problems 3.5.2 and 3.5.3: A tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (\(R\)), four yellow (\(Y\)), and five blue (\(B\)) beads. For the coin, \(P(H) = \frac{2}{3}\) and \(P(T) = \frac{1}{3}\), where \(H\) is heads and \(T\) is tails. The bead branches carry \(P(R) = \frac{3}{12}\), \(P(Y) = \frac{4}{12}\), and \(P(B) = \frac{5}{12}\) after either coin result.
Problem 1 Solution
Step 1 — name the two stages. The first stage is whether the man develops cancer, so it forks into \(C\) and its complement \(C'\). The second stage is whether he gets at least one false positive, so each first-stage branch forks again into \(F\) and \(F'\).
Step 2 — put the given rates on the first set of branches.
$$P(C) = 0.4567 \qquad P(C') = 1 - 0.4567 = 0.5433$$Step 3 — put the false-positive rates on the second set. The problem gives one false-positive rate for men in general and gives no reason to think it changes with cancer status, so both second-stage forks carry the same pair:
$$P(F) = 0.51 \qquad P(F') = 1 - 0.51 = 0.49$$Step 4 — multiply along each path.
| Stage 1 | Stage 2 | Outcome | Path probability |
|---|---|---|---|
| \(C\): 0.4567 | \(F\): 0.51 | \(C\) AND \(F\) | 0.2329 |
| \(C\): 0.4567 | \(F'\): 0.49 | \(C\) AND \(F'\) | 0.2238 |
| \(C'\): 0.5433 | \(F\): 0.51 | \(C'\) AND \(F\) | 0.2771 |
| \(C'\): 0.5433 | \(F'\): 0.49 | \(C'\) AND \(F'\) | 0.2662 |
Step 5 — check the tree. The four path probabilities add to \(0.2329 + 0.2238 + 0.2771 + 0.2662 = 1.0000\), which is what a complete tree always does.
Answer: a two-stage tree with first branches \(C = 0.4567\) and \(C' = 0.5433\), each splitting into \(F = 0.51\) and \(F' = 0.49\), and the four tip values 0.2329, 0.2238, 0.2771, and 0.2662.
Problem 2. Find the probability of tossing a head on the coin AND drawing a red bead, that is \(P(H \text{ AND } R)\).
a) \(\frac{2}{3}\)
b) \(\frac{5}{15}\)
c) \(\frac{6}{36}\)
d) \(\frac{5}{36}\)
Problem 2 Solution
Step 1 — this names one complete path, heads on the coin and then a red bead, so multiply the two branches.
Step 2 — multiply.
$$P(H \text{ AND } R) = \left(\frac{2}{3}\right)\left(\frac{3}{12}\right) = \frac{6}{36} = \frac{1}{6}$$Step 3 — check the offered choices. Choice a is just \(P(H)\) and ignores the bead. Choice b has a denominator that appears nowhere in this experiment. Choice d is the head-and-blue path, not head-and-red.
Answer: c, \(\frac{6}{36}\).
Problem 3. Find \(P(\text{blue bead})\).
a) \(\frac{15}{36}\)
b) \(\frac{10}{36}\)
c) \(\frac{10}{12}\)
d) \(\frac{6}{36}\)
Problem 3 Solution
Step 1 — find every path that ends in a blue bead. There are two: heads then blue, and tails then blue.
Step 2 — multiply along each one.
$$P(H \text{ AND } B) = \left(\frac{2}{3}\right)\left(\frac{5}{12}\right) = \frac{10}{36} \qquad P(T \text{ AND } B) = \left(\frac{1}{3}\right)\left(\frac{5}{12}\right) = \frac{5}{36}$$Step 3 — add across the paths.
$$P(B) = \frac{10}{36} + \frac{5}{36} = \frac{15}{36} = \frac{5}{12}$$The answer simplifies back to \(\frac{5}{12}\), which is exactly the share of blue beads in the cup. That makes sense: the coin decides nothing about the beads, so the unfair coin cannot change the chance of drawing blue.
Answer: a, \(\frac{15}{36}\).
Problem 4. A box of cookies contains three chocolate and seven butter cookies. Mei randomly selects a cookie and eats it. Then she randomly selects another cookie and eats it.
a) Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
b) Are the probabilities for the flavor of the SECOND cookie that Mei selects independent of her first selection? Explain.
c) For each complete path through the tree, write the event it represents and find the probabilities.
d) Let \(S\) be the event that both cookies selected were the same flavor. Find \(P(S)\).
e) Let \(D\) be the event that the cookies selected were different flavors. Find \(P(D)\) by two different methods: by using the complement rule, and by using the branches of the tree. Your answers should be the same with both methods.
f) Let \(B\) be the event that the second cookie selected is a butter cookie. Find \(P(B)\).
Problems 3.5.5 and 3.5.6: Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled. Let \(G_1\) = first card is green and \(G_2\) = second card is green.
Problem 4 Solution
Part a — build the tree. The box holds \(3 + 7 = 10\) cookies. Mei eats the first one, so only 9 remain for the second selection. Writing \(C\) for chocolate and \(B\) for butter:
| First cookie | Second cookie | Outcome | Path probability |
|---|---|---|---|
| \(C\): \(\frac{3}{10}\) | \(C\): \(\frac{2}{9}\) | CC | \(\frac{6}{90}\) |
| \(C\): \(\frac{3}{10}\) | \(B\): \(\frac{7}{9}\) | CB | \(\frac{21}{90}\) |
| \(B\): \(\frac{7}{10}\) | \(C\): \(\frac{3}{9}\) | BC | \(\frac{21}{90}\) |
| \(B\): \(\frac{7}{10}\) | \(B\): \(\frac{6}{9}\) | BB | \(\frac{42}{90}\) |
Part b — no, they are not independent. She ate the first cookie, so it is gone. If the first was chocolate, only 2 of the remaining 9 are chocolate; if the first was butter, 3 of the remaining 9 are. The second probability changes with the first result, which is the definition of dependent.
Part c — the four paths and what they mean. CC = both chocolate, \(\frac{6}{90}\); CB = chocolate then butter, \(\frac{21}{90}\); BC = butter then chocolate, \(\frac{21}{90}\); BB = both butter, \(\frac{42}{90}\). They add to \(\frac{90}{90} = 1\).
Part d — same flavor means CC or BB.
$$P(S) = \frac{6}{90} + \frac{42}{90} = \frac{48}{90} = \frac{8}{15}$$Part e — different flavors, two ways. By the complement rule:
$$P(D) = 1 - P(S) = 1 - \frac{48}{90} = \frac{42}{90} = \frac{7}{15}$$By the branches, adding the two mixed paths:
$$P(D) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90} = \frac{7}{15}$$Both methods give \(\frac{7}{15}\), as they must.
Part f — the second cookie is butter on the CB and BB paths.
$$P(B \text{ second}) = \frac{21}{90} + \frac{42}{90} = \frac{63}{90} = \frac{7}{10}$$Notice this equals the plain share of butter cookies in the original box. Before you know anything about the first pick, the second cookie is just as likely to be butter as the first one was.
Answer: a. the tree above; b. no, the pool shrinks; c. \(\frac{6}{90}, \frac{21}{90}, \frac{21}{90}, \frac{42}{90}\); d. \(\frac{8}{15}\); e. \(\frac{7}{15}\) both ways; f. \(\frac{7}{10}\).
Problem 5. Suppose that you randomly draw two cards, one at a time, with replacement.
a) Draw a tree diagram of the situation.
b) Find \(P(G_1 \text{ AND } G_2)\).
c) Find \(P(\text{at least one green})\).
d) Find \(P(G_2 \mid G_1)\).
e) Are \(G_2\) and \(G_1\) independent events? Explain why or why not.
Problem 5 Solution
Part a — build the tree with replacement. Eight cards, five green (\(G\)) and three yellow (\(Y\)). The first card goes back, so both stages carry the same branch probabilities \(\frac{5}{8}\) and \(\frac{3}{8}\).
| First card | Second card | Outcome | Path probability |
|---|---|---|---|
| \(G\): \(\frac{5}{8}\) | \(G\): \(\frac{5}{8}\) | GG | \(\frac{25}{64}\) |
| \(G\): \(\frac{5}{8}\) | \(Y\): \(\frac{3}{8}\) | GY | \(\frac{15}{64}\) |
| \(Y\): \(\frac{3}{8}\) | \(G\): \(\frac{5}{8}\) | YG | \(\frac{15}{64}\) |
| \(Y\): \(\frac{3}{8}\) | \(Y\): \(\frac{3}{8}\) | YY | \(\frac{9}{64}\) |
Part b — multiply along the GG path.
$$P(G_1 \text{ AND } G_2) = \left(\frac{5}{8}\right)\left(\frac{5}{8}\right) = \frac{25}{64}$$Part c — go through the complement. "At least one green" is everything except "no green at all", and the only path with no green is YY:
$$P(\text{at least one green}) = 1 - \frac{9}{64} = \frac{55}{64}$$Part d — read the second-stage branch. The card was replaced, so the deck is untouched:
$$P(G_2 \mid G_1) = \frac{5}{8}$$Part e — yes, they are independent. \(P(G_2 \mid G_1) = \frac{5}{8}\) and \(P(G_2) = \frac{5}{8}\) are equal, so knowing the first card was green tells you nothing about the second. Replacement is what makes this true.
Answer: a. the tree above; b. \(\frac{25}{64}\); c. \(\frac{55}{64}\); d. \(\frac{5}{8}\); e. yes, because \(P(G_2 \mid G_1) = P(G_2)\).
Problem 6. Suppose that you randomly draw two cards, one at a time, without replacement.
a) Draw a tree diagram of the situation.
b) Find \(P(G_1 \text{ AND } G_2)\).
c) Find \(P(\text{at least one green})\).
d) Find \(P(G_2 \mid G_1)\).
e) Are \(G_2\) and \(G_1\) independent events? Explain why or why not.
Problem 6 Solution
Part a — build the tree without replacement. The first card stays out, so every second-stage denominator drops from 8 to 7, and the green count drops by one on the branch where a green was taken.
| First card | Second card | Outcome | Path probability |
|---|---|---|---|
| \(G\): \(\frac{5}{8}\) | \(G\): \(\frac{4}{7}\) | GG | \(\frac{20}{56}\) |
| \(G\): \(\frac{5}{8}\) | \(Y\): \(\frac{3}{7}\) | GY | \(\frac{15}{56}\) |
| \(Y\): \(\frac{3}{8}\) | \(G\): \(\frac{5}{7}\) | YG | \(\frac{15}{56}\) |
| \(Y\): \(\frac{3}{8}\) | \(Y\): \(\frac{2}{7}\) | YY | \(\frac{6}{56}\) |
Part b — multiply along the GG path.
$$P(G_1 \text{ AND } G_2) = \left(\frac{5}{8}\right)\left(\frac{4}{7}\right) = \frac{20}{56} = \frac{5}{14}$$Part c — complement again. The only path with no green is YY, worth \(\frac{6}{56}\):
$$P(\text{at least one green}) = 1 - \frac{6}{56} = \frac{50}{56} = \frac{25}{28}$$Part d — read the second-stage branch under a green first card. One green is gone, so four greens remain among seven cards:
$$P(G_2 \mid G_1) = \frac{4}{7}$$Part e — no, they are not independent. \(P(G_2 \mid G_1) = \frac{4}{7} \approx 0.571\) but \(P(G_2) = \frac{5}{8} = 0.625\). Drawing a green first lowers the chance of green second, so the two events depend on each other. Comparing with Problem 3.5.5, the only thing that changed was putting the card back.
Answer: a. the tree above; b. \(\frac{5}{14}\); c. \(\frac{25}{28}\); d. \(\frac{4}{7}\); e. no, because \(\frac{4}{7} \neq \frac{5}{8}\).
Problem 7. When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one-Euro coin was a fair coin. They spun the coin rather than tossing it, and found that out of 250 spins, 140 showed a head (event \(H\)) while 110 showed a tail (event \(T\)). On that basis they claimed that it is not a fair coin.
a) Based on the given data, find \(P(H)\) and \(P(T)\).
b) Use a tree to find the probabilities of each possible outcome for the experiment of spinning the coin twice.
c) Use the tree to find the probability of obtaining exactly one head in two spins of the coin.
d) Use the tree to find the probability of obtaining at least one head.
Problem 7 Solution
Part a — turn the spin counts into probabilities. Out of 250 spins, 140 were heads and 110 were tails:
$$P(H) = \frac{140}{250} = 0.56 \qquad P(T) = \frac{110}{250} = 0.44$$A fair coin would give 0.5 each, so this coin leans toward heads — which is the professors' claim.
Part b — build the two-spin tree. Each spin is a fresh spin of the same coin, so both stages carry 0.56 and 0.44.
| First spin | Second spin | Outcome | Path probability |
|---|---|---|---|
| \(H\): 0.56 | \(H\): 0.56 | HH | 0.3136 |
| \(H\): 0.56 | \(T\): 0.44 | HT | 0.2464 |
| \(T\): 0.44 | \(H\): 0.56 | TH | 0.2464 |
| \(T\): 0.44 | \(T\): 0.44 | TT | 0.1936 |
The four values add to 1.0000, so the tree is complete.
Part c — exactly one head is HT or TH, so add those two paths.
$$P(\text{exactly one head}) = 0.2464 + 0.2464 = 0.4928$$Part d — at least one head is everything but TT.
$$P(\text{at least one head}) = 1 - 0.1936 = 0.8064$$You could also add the three paths HH, HT, and TH: \(0.3136 + 0.2464 + 0.2464 = 0.8064\).
Answer: a. \(P(H) = 0.56\), \(P(T) = 0.44\); b. HH 0.3136, HT 0.2464, TH 0.2464, TT 0.1936; c. 0.4928; d. 0.8064.
Key Terms
tree diagram — a graph of a multi-stage experiment in which each stage is a set of branches and each complete path from trunk to tip is one outcome.
branch — a single labeled edge of a tree diagram, carrying the frequency or probability of one step of the experiment.
path — a complete route through a tree diagram from the first stage to the last; its probability is the product of its branches.
with replacement — sampling in which the selected object is returned before the next selection, so the pool and its probabilities are unchanged and the stages are independent.
without replacement — sampling in which the selected object is kept out, so the pool shrinks and the second-stage probabilities depend on the first.
Venn diagram — a picture of an experiment consisting of a box for the sample space \(S\) with circles or ovals inside it for the events.