13.4 Cardinality

Often times we are interested in the number of items in a set or subset. This is called the cardinality of the set.

Sometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying.

Example 3

A survey asks 200 people “What beverage do you drink in the morning”, and offers choices:

Tea only
Coffee only
Both coffee and tea

Suppose 20 report tea only, 80 report coffee only, 40 report both. How many people drink tea in the morning? How many people drink neither tea or coffee?

Show solution

Venn diagram of two overlapping circles inside a rectangle, Coffee on the left and Tea on the right, holding the results of a 200 person survey. The Coffee only region is marked 80, the overlap is marked 40 and the Tea only region is marked 20; nothing is written in the area outside both circles. This question can most easily be answered by creating a Venn diagram. We can see that we can find the people who drink tea by adding those who drink only tea to those who drink both: 60 people.

We can also see that those who drink neither are those not contained in the any of the three other groupings, so we can count those by subtracting from the cardinality of the universal set, 200.

$200-20-80-40=60$ people who drink neither.

Example 4

A survey asks: Which online services have you used in the last month:

Twitter
Facebook
Have used both

The results show 40% of those surveyed have used Twitter, 70% have used Facebook, and 20% have used both. How many people have used neither Twitter or Facebook?

Show solution

Let $T$ be the set of all people who have used Twitter, and $F$ be the set of all people who have used Facebook. Notice that while the cardinality of $F$ is $70 \%$ and the cardinality of $T$ is $40 \%$ , the cardinality of $F \cup T$ is not simply $70 \%+40 \%$ , since that would count those who use both services twice. To find the cardinality of $F \cup T$ , we can add the cardinality of $F$ and the cardinality of $T$ , then subtract those in intersection that we've counted twice. In symbols,

# The book's survey: 40% used Twitter, 70% used Facebook, 20% used both.
# EDIT these four numbers and re-run.
nT, nF, nBoth, nU = 40, 70, 20, 100

nUnion = nF + nT - nBoth
print(f"n(T)        = {nT}")
print(f"n(F)        = {nF}")
print(f"n(F inter T)= {nBoth}")
print(f"n(F union T)= n(F) + n(T) - n(F inter T) = {nF} + {nT} - {nBoth} = {nUnion}")
print(f"n((F union T)c) = n(U) - n(F union T) = {nU} - {nUnion} = {nU - nUnion}")
print(f"\nnaive answer without subtracting the overlap: {nF + nT}"
      f"  -> {nF + nT - nUnion} people counted twice")

# Proof that this is not a trick: build actual people and count them.
both = set(range(nBoth))
T = both | set(range(100, 100 + nT - nBoth))
F = both | set(range(200, 200 + nF - nBoth))
print("\nsame thing with real sets of people:")
print(f"   len(T) = {len(T)}, len(F) = {len(F)}, len(T & F) = {len(T & F)}")
print(f"   len(T | F) = {len(T | F)}   formula said {nUnion}   match? {len(T | F) == nUnion}")

if nBoth > min(nT, nF):
    print("\nWARNING: more people use both than use one of them - impossible survey data.")
if nUnion > nU:
    print("\nWARNING: the union is bigger than the universal set - impossible survey data.")
# Try it: set nBoth = 0 (nobody uses both) and watch the subtraction vanish.

$\mathrm{n}(F \cup T)=\mathrm{n}(F)+\mathrm{n}(T)-\mathrm{n}(F \cap T)$

$\mathrm{n}(F \cup T)=70 \%+40 \%-20 \%=90 \%$

Now, to find how many people have not used either service, we're looking for the cardinality of $(F \cup T)^{c}$ . since the universal set contains $100 \%$ of people and the cardinality of $F \cup T=90 \%$ , the cardinality of $(F \cup 7)^{c}$ must be the other $10 \%$

The previous example illustrated two important properties

Note

Cardinality properties

$\mathrm{n}(A \cup B)=\mathrm{n}(A)+\mathrm{n}(B)-\mathrm{n}(A \cap B)$

$n\left(A^{\circ}\right)=n(U)-n(A)$

Notice that the first property can also be written in an equivalent form by solving for the cardinality of the intersection:

$\mathrm{n}(A \cap B)=\mathrm{n}(A)+\mathrm{n}(B)-\mathrm{n}(A \cup B)$

Example 5

Add text here.Fifty students were surveyed, and asked if they were taking a social science (SS), humanities (HM) or a natural science (NS) course the next quarter.

import matplotlib.pyplot as plt

# The book's survey data. EDIT any number and re-run.
TOTAL = 50
SS, HM, NS = 21, 26, 19          # taking each subject
SS_HM, SS_NS, HM_NS = 9, 7, 10   # taking each pair (includes the all-three students)
ALL3 = 3

only_ss_hm = SS_HM - ALL3
only_ss_ns = SS_NS - ALL3
only_hm_ns = HM_NS - ALL3
only_ss = SS - only_ss_hm - only_ss_ns - ALL3
only_hm = HM - only_ss_hm - only_hm_ns - ALL3
only_ns = NS - only_ss_ns - only_hm_ns - ALL3
inside = (only_ss + only_hm + only_ns + only_ss_hm + only_ss_ns + only_hm_ns + ALL3)
none = TOTAL - inside

print("working from the middle of the diagram outwards")
print(f"   all three                      {ALL3}")
print(f"   SS and HM but not NS           {SS_HM} - {ALL3} = {only_ss_hm}")
print(f"   SS and NS but not HM           {SS_NS} - {ALL3} = {only_ss_ns}")
print(f"   HM and NS but not SS           {HM_NS} - {ALL3} = {only_hm_ns}")
print(f"   SS only    {SS} - {only_ss_hm} - {only_ss_ns} - {ALL3} = {only_ss}")
print(f"   HM only    {HM} - {only_ss_hm} - {only_hm_ns} - {ALL3} = {only_hm}")
print(f"   NS only    {NS} - {only_ss_ns} - {only_hm_ns} - {ALL3} = {only_ns}")
print(f"   at least one course            {inside}")
print(f"   no course                      {TOTAL} - {inside} = {none}")
if min(only_ss, only_hm, only_ns, only_ss_hm, only_ss_ns, only_hm_ns, none) < 0:
    print("   WARNING: a region came out negative - the survey numbers are inconsistent.")

fig, ax = plt.subplots(figsize=(5.6, 5.4))
spots = [(-0.62, 0.42, "SS", only_ss, -1.35, 0.95), (0.62, 0.42, "HM", only_hm, 1.35, 0.95),
         (0.0, -0.62, "NS", only_ns, 0.0, -1.55)]
for cx, cy, lab, val, lx, ly in spots:
    ax.add_patch(plt.Circle((cx, cy), 1.15, alpha=0.25, lw=2, edgecolor="black"))
    ax.text(lx, ly, f"{lab}\n{val}", ha="center", fontsize=12)
for tx, ty, val in ((0.0, 0.66, only_ss_hm), (-0.5, -0.28, only_ss_ns),
                    (0.5, -0.28, only_hm_ns), (0.0, 0.05, ALL3)):
    ax.text(tx, ty, str(val), ha="center", fontsize=12)
ax.text(-1.95, -1.85, f"none: {none}", fontsize=11)
ax.set_xlim(-2.1, 2.1); ax.set_ylim(-2.1, 2.0)
ax.set_aspect("equal"); ax.axis("off")
ax.set_title(f"{TOTAL} students surveyed", fontsize=13)
plt.show()

$\begin{array}{ll} \text{21 were taking a SS course} & \text{26 were taking a HM course} \\ \text{19 were taking a NS course} & \text{9 were taking SS and HM} \\ \text{7 were taking SS and NS} & \text{10 were taking HM and NS} \\ \text{3 were taking all three} & \text{7 were taking none} \end{array}$

How many students are only taking a SS course?

Show solution

Venn diagram of three overlapping circles inside a rectangle, SS at the upper left, HM at the upper right and NS below, with all eight regions given letter names: a is SS only, b is SS and HM only, c is HM only, d is SS and NS only, e is the overlap of all three, f is HM and NS only, g is NS only, and h is the area outside all three circles. It might help to look at a Venn diagram.

From the given data, we know that there are 3 students in region $e$ and 7 students in region $h$

since 7 students were taking a $S S$ and $N S$ course, we know that $n(d)+n(e)=7$ . since we know there are 3 students in region 3, there must be
$7-3=4$ students in region $d$

Similarly, since there are 10 students taking $\mathrm{HM}$ and $\mathrm{NS}$ , which includes regions $e$ and $f$ , there must be

$10-3=7$ students in region $f$

Since 9 students were taking $\mathrm{SS}$ and $\mathrm{HM}$ , there must be $9-3=6$ students in region $b$

Now, we know that 21 students were taking a SS course. This includes students from regions $a, b, d,$ and $e.$ since we know the number of students in all but region $a,$ we can determine that $21-6-4-3=8$ students are in region $a$

8 students are taking only a SS course.

Your Turn

Try it Now 4

One hundred fifty people were surveyed and asked if they believed in UFOs, ghosts, and Bigfoot.

$\begin{array}{ll} \text{43 believed in UFOs} & \text{44 believed in ghosts} \\ \text{25 believed in Bigfoot} & \text{10 believed in UFOs and ghosts} \\ \text{8 believed in ghosts and Bigfoot} & \text{5 believed in UFOs and Bigfoot} \\ \text{2 believed in all three} & \text{} \end{array}$

How many people surveyed believed in at least one of these things?

Answer

Venn diagram of three overlapping circles inside a rectangle, UFOs at the upper left, Ghosts at the upper right and Bigfoot below, filled in for a survey of 150 people: 30 in UFOs only, 8 in UFOs and Ghosts only, 28 in Ghosts only, 3 in UFOs and Bigfoot only, 2 in all three, 6 in Ghosts and Bigfoot only, 14 in Bigfoot only, and 59 outside all three circles. Starting with the intersection of all three circles, we work our way out. since 10 people believe in UFOs and Ghosts, and 2 believe in all three, that leaves 8 that believe in only UFOs and Ghosts. We work our way out, filling in all the regions. Once we have, we can add up all those regions, getting 91 people in the union of all three sets. This leaves $150-91=59$ who believe in none.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.