Solved Examples on Venn Diagrams for Three or More Sets

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Corequisites: Probability

Solve all questions
Draw Venn Diagrams for each question and answer as applicable
Show all work.

Reference Venn Diagram: We shall use it for some questions.
Working Copy - 3 Sets

$ \mu = \{1, 2, 3, 4, 5, 6, 7, 8\} \\[3ex] A = \{1, 4, 5, 7\} \\[3ex] A' = \{2, 3, 6, 8\} \\[3ex] B = \{2, 5, 6, 7\} \\[3ex] B' = \{1, 3, 4, 8\} \\[3ex] C = \{3, 4, 6, 7\} \\[3ex] C' = \{1, 2, 5, 8\} $

(1.) NSC Complaints about a restaurant fell into three main categories: the menu (M), the food (F) and the service (S).
In total 173 complaints were received in a certain month.
The complaints were as follows:

110 complained about the menu.
55 complained about the food.
67 complained about the service.
20 complained about the menu and the food, but not the service.
11 complained about the menu and the service, but not the food.
16 complained about the food and the service, but not the menu.
The number who complained about all three is unknown.

(1.1) Draw a Venn Diagram to illustrate the above information.

(1.2) Determine the number of people who complained about ALL THREE categories.

(1.3) Determine the probability that a complaint selected at random from those received, complained about AT LEAST TWO of the categories (that is. menu, food and service).

Let the number who complained about all three = $x$

$ n(\mu) = 173 \\[3ex] n(M) = 110 \\[3ex] n(F) = 55 \\[3ex] n(S) = 67 \\[3ex] n(M\;\;and\;\;F\;\;only) = 20 \\[3ex] n(M\;\;and\;\;S\;\;only) = 11 \\[3ex] n(F\;\;and\;\;S\;\;only) = 16 \\[3ex] n(M\;\;only) = 110 - (20 + x + 11) \\[3ex] n(F\;\;only) = 55 - (20 + x + 16) \\[3ex] n(S\;\;only) = 67 - (11 + x + 16) \\[3ex] (1.1) \\[3ex] $ The Venn Diagram is drawn as shown:

$ (1.2) \\[3ex] 110 - (20 + x + 11) + 55 - (20 + x + 16) + 67 - (11 + x + 16) + 20 + 11 + 16 + x = 173 \\[3ex] 110 - (31 + x) + 55 - (36 + x) + 67 - (27 + x) + 47 + x = 173 \\[3ex] 110 - 31 - x + 55 - 36 - x + 67 - 27 - x + 47 + x = 173 \\[3ex] 185 - 2x = 173 \\[3ex] 185 - 173 = 2x \\[3ex] 12 = 2x \\[3ex] 2x = 12 \\[3ex] x = \dfrac{12}{2} \\[5ex] x = 6 \\[3ex] $ $6$ people complained about all three categories.

Let $E$ be the event of selecting a complaint about at least two of the categories
At least two means two or more
It means:
$M$ and $F$ only ($20$) or
$M$ and $S$ only ($11$) or
$F$ and $S$ only ($16$) or
$M$, $F$, and $S$ ... in other words all three = $6$

$ (1.3) \\[3ex] n(E) = 20 + 11 + 16 + 6 \\[3ex] n(E) = 53 \\[3ex] n(S) = n(\mu) = 173 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{53}{173} $

(2.) JAMB

The Venn diagram above shows a class of 40 students with the games they play.
How many of the students play two games only?

$ A.\;\; 19 \\[3ex] B.\;\; 16 \\[3ex] C.\;\; 15 \\[3ex] D.\;\; 4 \\[3ex] $

The number of studnets who play two games only are:
those who play only Football and Tennis : $5$ and
those who play only Football and Athletics : $7$ and
those who play only Tennis and Athletics : $3$

$ n(Students\;\;who\;\;play\;\;only\;\;two\;\;games) \\[3ex] = 5 + 7 + 3 \\[3ex] = 15\;\;students \\[3ex] $ $15$ students play only two games

(3.) NSC A survey is conducted among $174$ students.
The results are shown below.

37 study Life Sciences
60 study Physical Sciences
111 study Mathematics
29 study Life Sciences and Mathematics
50 study Mathematics and Physical Sciences
13 study Physical Sciences and Life Sciences
45 do not study any of Life Sciences, Mathematics or Physical Sciences
x students study Life Sciences, Mathematics and Physical Sciences

(3.1) Draw a Venn diagram to represent the information above.
(3.2) Show that x = 13
(3.3) If a student were selected at random, calculate the probabability that he studies the following:
(3.3.1) Mathematics and Physical Sciences but not Life Sciences
(3.3.2) Only one of Mathematics or Physical Sciences or Life Sciences

$ Let: \\[3ex] Life\;\;Sciences = L \\[3ex] Physical\;\;Sciences = P \\[3ex] Mathematics = M \\[3ex] (3.1) \\[3ex] $ The Venn Diagram is drawn as shown:

$ (3.2) \\[3ex] 37 − [(13 − 𝑥) + 𝑥 + (29 − 𝑥)] + 60 − [(13 − 𝑥) + 𝑥 + (50 − 𝑥)] + 111 − [(29 − 𝑥) + 𝑥 + (50 − 𝑥)] + 13 - x + 29 - x + 50 - x + x + 45 = 174 \\[3ex] 37 - [13 - x + x + 29 - x] + 60 - [13 - x + x + 50 - x] + 111 - [29 - x + x + 50 - x] + 137 - 2x = 174 \\[3ex] 37 - [42 - x] + 60 - [63 - x] + 111 - [79 - x] + 137 - 2x = 174 \\[3ex] 37 - 42 + x + 60 - 63 + x + 111 - 79 + x + 137 - 2x = 174 \\[3ex] 161 + x = 174 \\[3ex] x = 174 - 161 \\[3ex] x = 13 \\[3ex] $ $13$ students study Life Sciences, Mathematics and Physical Sciences

Let $E$ be the event of selecting a student that studies Mathematics and Physical Sciences but not Life Sciences

$ (3.3.1) \\[3ex] Mathematics\;\;and\;\;Physical\;\;Sciences\;\;but\;\;not\;\;Life\;\;Sciences \\[3ex] = Mathematics\;\;and\;\;Physical\;\;Sciences\;\;only \\[3ex] = 50 - x \\[3ex] = 50 - 13 \\[3ex] = 37 \\[3ex] n(E) = 37 \\[3ex] n(S) = 174 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{37}{174} \\[5ex] $ Let $K$ be the event of selecting a student that studies only one of Mathematics or Physical Sciences or Life Sciences

$ (3.3.2) \\[3ex] Only\;\;one\;\;of\;\;Mathematics\;\;or\;\;Physical\;\;Sciences\;\;or\;\;Life\;\;Sciences \\[3ex] = only\;\;Mathematics\;\;or\;\;only\;\;Physical\;\;Sciences\;\;or\;\;only\;\;Life\;\;Sciences \\[3ex] Only\;\;Mathematics \\[3ex] = 37 − [(13 − 𝑥) + 𝑥 + (29 − 𝑥)] \\[3ex] = 37 - [13 - x + x + 29 - x] \\[3ex] = 37 - [42 - x] \\[3ex] = 37 - 42 + x \\[3ex] = -5 + x \\[3ex] = -5 + 13 \\[3ex] = 8 \\[3ex] Only\;\;Physical\;\;Sciences \\[3ex] = 60 − [(13 − 𝑥) + 𝑥 + (50 − 𝑥)] \\[3ex] = 60 - [13 - x + x + 50 - x] \\[3ex] = 60 - [63 - x] \\[3ex] = 60 - 63 + x \\[3ex] = -3 + x \\[3ex] = -3 + 13 \\[3ex] = 10 \\[3ex] Only\;\;Life\;\;Sciences \\[3ex] = 111 − [(29 − 𝑥) + 𝑥 + (50 − 𝑥)] \\[3ex] = 111 - [29 - x + x + 50 - x] \\[3ex] = 111 - [79 - x] \\[3ex] = 111 - 79 + x \\[3ex] = 32 + x \\[3ex] = 32 + 13 \\[3ex] = 45 \\[3ex] n(K) = 8 + 10 + 45 \\[3ex] n(K) = 63 \\[3ex] P(K) = \dfrac{n(K)}{n(S)} \\[5ex] P(K) = \dfrac{63}{174} \\[5ex] P(K) = \dfrac{21}{58} $

(4.) ACT In a large high school, some teachers teach only 1 subject, and some teachers more than 1 subject.
Using the information given in the table below about the math, science, and gym teachers in the school, how many teachers teach math only?

Number of teachers	Subject(s) taught
12 10 20 6 5 2 1	at least 1 class of math at least 1 class of gym at least 1 class of science both gym and science but not math both math and science but not gym gym only math, gym, and science

$ F.\;\; 1 \\[3ex] G.\;\; 2 \\[3ex] H.\;\; 5 \\[3ex] J.\;\; 16 \\[3ex] K.\;\; 28 \\[3ex] $

$ Let: \\[3ex] math = M \\[3ex] gym = G \\[3ex] science = C \\[3ex] n(M) = 12 \\[3ex] n(G) = 10 \\[3ex] n(C) = 20 \\[3ex] n(G\;\;and\;\;C\;\;ONLY) = n(G \cap C \cap M') = 6 \\[3ex] n(M\;\;and\;\;C\;\;ONLY) = n(M \cap C \cap G') = 5 \\[3ex] n(G\;\;ONLY) = n(G \cap M' \cap C') = 2 \\[3ex] n(M \cap G \cap C) = 1 \\[3ex] n(M\;\;and\;\;G\;\;ONLY) = n(M \cap G \cap C') = r \\[3ex] n(M\;\;ONLY) = n(M \cap G' \cap C') = p \\[3ex] $

$ r + 2 + 1 + 6 = 10 \\[3ex] r + 9 = 10 \\[3ex] r = 10 - 9 \\[3ex] r = 1 \\[3ex] p + r + 5 + 1 = 12 \\[3ex] p + 1 + 6 = 12 \\[3ex] p + 7 = 12 \\[3ex] p = 12 - 7 \\[3ex] p = 5 \\[3ex] $ 5 teachers teach only math.

(5.) WASSCE A selection interview was conducted for 110 students into various subjects in a Science department.
25 were selected for Physics, 45 for Biology, and 48 for Mathematics.
10 were selected for Physics and Mathematics, 8 for Biology and Mathematics, and 6 for Physics and Biology, while 5 were selected for all three subjects.
(i.) Illustrate the information on a Venn diagram
(ii.) How many students were selected for Biology but neither Physics nor Mathematics?
(iii.) How many students were not selected for any of the three subjects?

$ Let: \\[3ex] Physics = P \\[3ex] Biology = B \\[3ex] Mathematics = M \\[3ex] n(P) = 25 \\[3ex] n(B) = 45 \\[3ex] n(M) = 48 \\[3ex] n(P \cap M) = 10 \\[3ex] n(B \cap M) = 8 \\[3ex] n(P \cap B) = 6 \\[3ex] n(P \cap B \cap M) = 5 \\[3ex] n(\mu) = 110 \\[3ex] $ (i.)
The Venn diagram is:

$ (ii.) \\[3ex] n(B\;\;ONLY) = n(B \cap P' \cap M') = 36 \\[3ex] $ 36 students were selected for Biology only

$ (iii) \\[3ex] n(neither) = n(P' \cap B' \cap M') = k \\[3ex] \implies \\[3ex] 14 + 36 + 35 + 5 + 1 + 3 + 5 + k = 110 \\[3ex] 99 + k = 110 \\[3ex] k = 110 - 99 \\[3ex] k = 11 \\[3ex] $ 11 students were not selected for any of the three subjects

(6.) CSEC The universal set U = {b, d, e, f, g, i, k, s, t, v, w}
The Venn diagram below shows U and three sets, M, P and R, which are subsets of U

(i) State the value of $n(P \cup R)$

(ii) List the members of
(a.) $M \cap P$
(b.) $M \cup R'$

$ M = \{b, d, i, k\} \\[3ex] P = \{b, d, e, f, s, v\} \\[3ex] R = \{d, e, f, g, i\} \\[3ex] \mu = \{b, d, e, f, g, i, k, s, t, v, w\} \\[3ex] (i) \\[3ex] P \cup R = \{b, d, e, f, g, i, s, v\} \\[3ex] n(P \cup R) = 8 \\[3ex] (ii) \\[3ex] (a.) \\[3ex] M \cap P = \{b, d\} \\[3ex] (b.) \\[3ex] R' = \{b, k, s, t, v, w\} \\[3ex] M \cup R' = \{b, d, i, k, s, t, v, w\} $

(7.) Hebrews 13:8 School of Thought offers Algebra, Biology, and Chemistry courses.
Six students are enrolled in all three courses.
Twenty one students are enrolled in Algebra and Biology.
Twenty six students are enrolled in Algebra and Chemistry.
Seven students are enrolled in Biology and Chemistry.
Forty students are enrolled in Chemistry.
Among those enrolled in Algebra, the number of students studying only Algebra and only Algebra and Biology is twenty.
Among those enrolled in Bology, the number of students studying only Biology and only Algebra and Biology is twenty five.
Two students are not enrolled in any of the three courses.

(a.) Represent this information on a Venn Diagram.

(b.) Complete the Venn Diagram.

$ Let: \\[3ex] Algebra = A \\[3ex] Biology = B \\[3ex] Chemistry = C \\[3ex] n(A \cap B \cap C) = 6 \\[3ex] n(A \cap B) = 21 \\[3ex] n(B \cap C) = 7 \\[3ex] n(C) = 40 \\[3ex] n(A \cap B' \cap C') + n(A \cap B \cap C') = 20 \\[3ex] But:\;\; \\[3ex] n(A \cap B' \cap C') + n(A \cap B \cap C') = n(A \cap C') \\[3ex] \implies n(A \cap C') = 20 \\[3ex] \implies n(A \cap B' \cap C') + n(A \cap B \cap C') = 20 \\[3ex] n(A' \cap B \cap C') + n(A \cap B \cap C') = 25 \\[3ex] But:\;\; \\[3ex] n(A' \cap B \cap C') + n(A \cap B \cap C') = n(B \cap C') \\[3ex] \implies n(B \cap C') = 25 \\[3ex] \implies n(A' \cap B \cap C') + n(A \cap B \cap C') = 25 \\[3ex] n(A' \cap B' \cap C') = 2 \\[3ex] $ Number 7-1

$ \underline{Only\;\;A\;\;and\;\;B} \\[3ex] n(A \cap B \cap C') \\[3ex] = 21 - 6 \\[3ex] = 15\;\;students \\[3ex] \underline{Only\;\;A\;\;and\;\;C} \\[3ex] n(A \cap B' \cap C) \\[3ex] = 26 - 6 \\[3ex] = 20\;\;students \\[3ex] \underline{Only\;\;B\;\;and\;\;C} \\[3ex] n(A' \cap B \cap C) \\[3ex] = 7 - 6 \\[3ex] = 1\;\;student \\[3ex] \underline{Only\;\;C} \\[3ex] n(A' \cap B' \cap C) \\[3ex] = 40 - (20 + 6 + 1) \\[3ex] = 40 - 27 \\[3ex] = 13\;\;students \\[3ex] \underline{Only\;\;A} \\[3ex] n(A \cap B' \cap C') + 15 = 20 \\[3ex] n(A \cap B' \cap C') \\[3ex] = 20 - 15 \\[3ex] = 5\;\;students \\[3ex] \underline{Only\;\;B} \\[3ex] n(A' \cap B \cap C') + 15 = 25 \\[3ex] n(A' \cap B \cap C') \\[3ex] = 25 - 15 \\[3ex] = 10\;\;students \\[3ex] \underline{Neither\;\;A\;\;nor\;\;B\;\;C} \\[3ex] n(A' \cap B' \cap C') = 2 \\[3ex] \underline{Universal\;\;Set} \\[3ex] n(\mu) \\[3ex] = 5 + 10 + 13 + 20 + 15 + 1 + 6 + 2 \\[3ex] = 72\;\;students \\[3ex] $ Number 7-2

(8.) GCSE $\mathcal{E}$ = {even numbers between 1 and 25}
A = {2, 8, 10, 14}
B = {6, 8, 20}
C = {8, 18, 20, 22}

(a) Complete the Venn diagram for this information.

A number is chosen at random from $\mathcal{E}$
(b) Find the probability that the number is a member of $A \cap B$

$ \mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] A = \{2, 8, 10, 14\} \\[3ex] B = \{6, 8, 20\} \\[3ex] C = \{8, 18, 20, 22\} \\[3ex] n(all\;\;three) = n(A \cap B \cap C) = \{8\} \\[3ex] A' = \{4, 6, 12, 16, 18, 20, 22, 24\} \\[3ex] B' = \{2, 4, 10, 12, 14, 16, 18, 22, 24\} \\[3ex] C' = \{2, 4, 6, 10, 12, 14, 16, 24\} \\[3ex] \underline{A\;\;and\;\;B\;\;only} \\[3ex] A \cap B = \{8\} \\[3ex] A \cap B \cap C' = \{\} \\[3ex] \underline{A\;\;and\;\;C\;\;only} \\[3ex] A \cap C = \{8\} \\[3ex] A \cap C \cap B' = \phi \\[3ex] \underline{B\;\;and\;\;C\;\;only} \\[3ex] B \cap C = \{8, 20\} \\[3ex] B \cap C \cap A' = \{20\} \\[3ex] \underline{A\;\;only} \\[3ex] B' \cap C' = \{2, 4, 10, 12, 14, 16, 24\} \\[3ex] A \cap B' \cap C' = \{2, 10, 14\} \\[3ex] \underline{B\;\;only} \\[3ex] A' \cap C' = \{4, 6, 12, 16, 24\} \\[3ex] B \cap A' \cap C' = \{6\} \\[3ex] \underline{C\;\;only} \\[3ex] A' \cap B' = \{4, 12, 16, 18, 22, 24\} \\[3ex] C \cap A' \cap B' = \{18, 22\} \\[3ex] $ (a) The Venn diagram is:

$ A \cap B = \{8\} \\[3ex] n(A \cap B) = 1 \\[3ex] \mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] n(\mathcal{E}) = 12 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{1}{12} $

(9.) WASSCE Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all three means of transport.
The number, x, of travellers who travelled by bus only was equal to the number who travelled by bus and car only.
If 47 people travelled by bus and 30 by train:

(a) Represent this information in a Venn diagram

(b) Calculate the
(i) value of x
(ii) number who travelled by at least two means

$ Let: \\[3ex] bus = B \\[3ex] train = T \\[3ex] car = C \\[3ex] $ (a) The Venn diagram is:

$ (b) \\[3ex] (i) \\[3ex] n(B) = 47 \\[3ex] \implies \\[3ex] x + x + 7 + 8 = 47 \\[3ex] 2x + 15 = 47 \\[3ex] 2x = 47 - 15 \\[3ex] 2x = 32 \\[3ex] x = \dfrac{32}{2} \\[5ex] x = 16 \\[3ex] (ii) \\[3ex] n(at\;\;least\;\;two\;\;means) \\[3ex] n(exactly\;\;two\;\;means) + n(exactly\;\;three\;\;means) \\[3ex] = n(Bus\;\;and\;\;Train\;\;only) \\[3ex] + n(Bus\;\;and\;\;Car\;\;only) \\[3ex] + n(Train\;\;and\;\;Car\;\;only) \\[3ex] + n(Bus\;\;and\;\;Train\;\;and\;\;Car) \\[3ex] = 7 + x + 3 + 8 \\[3ex] = 7 + 16 + 3 + 8 \\[3ex] = 34 \\[3ex] $ 34 people travelled by at least two means

(10.) ACT The counselors at Riverdale High School created the Venn diagram below to show the distribution of music elective placements of 150 students.
Students could request placement in 3 different music electives: Band, Orchestra, and Chorus.
What is the number of students who were placed in at least 2 of these music electives?

$ A.\;\; 5 \\[3ex] B.\;\; 13 \\[3ex] C.\;\; 16 \\[3ex] D.\;\; 22 \\[3ex] E.\;\; 27 \\[3ex] $

Let:
Band = B
Chorus = C
Orchestra = R

At least 2 means 2 or more
This means exactly 2 of the music electives and exactly 3 (all three) of the music electives.
The number of students who were placed in at least 2 of the musical electives are those placed in:
Band and Chorus ONLY
Band and Orchestra ONLY
Chrous and Orchestra ONLY
Band, Chorus, and Orchestra

$ n(at\;\;least\;2) = n(\ge 2) \\[3ex] = 13 + 7 + 2 + 5 \\[3ex] = 27 \\[3ex] $ 27 students were placed in at least two of the music electives.

(11.) The following numbers of patients in a (hypothetical) hospital on a single day were taking antibiotics (A), blood pressure medication (BP) and/or pain medication (P) in numbers summarized in the table below.


A only 10	A and BP only 17
BP only 5	A and P only 22
P only 26	BP and P only 17
None 5	All three 20

(a.) Draw a three-circle Venn diagram that summarizes the results in the table.
(b.) How many patients took pain medication or blood pressure medication?
(c.) How many patients took blood pressure medication but not antibiotics?
(d.) How many patients took (at least) blood pressure medication?
(e.) How many patients took only pain medication?
(f.) How many patients took exactly two medications?

(a.) The Venn diagram showing the information is:

(b.) Number of patients that took pain medication or blood pressure medication = $n(P \cup BP)$
= Medication + Blood pressure − (Medication and Blood pressure)

$ = n(P) + n(BP) - n(P \cup BP) \\[3ex] n(P) = 22 + 20 + 17 + 26 = 85 \\[3ex] n(BP) = 17 + 5 + 20 + 17 = 59 \\[3ex] n(P \cap BP) = 20 + 17 = 37 \\[3ex] \implies \\[3ex] n(P \cup BP) = 85 + 59 - 37 \\[3ex] n(P \cup BP) = 107 \\[3ex] $ 107 patients took pain medication or blood pressure medication

(c.) Number of patients that took blood pressure medication but not antibiotics = sum of: Antibiotics and Blood pressure medication only and Blood pressure medication only

$ = n(A \cap BP \cap P') + n(BP \cap A' \cap P') \\[3ex] = 17 + 5 \\[3ex] = 22 \\[3ex] $ 22 patients took blood pressure medication but not antibiotics

(d.) Number of patients that took (at least) blood pressure medication = sum of: Blood pressure medication only and Blood pressure medication and Antibiotics only and Blood pressure medication and Pain medication only and All three medications

$ = n(BP \cap P' \cap A') + n(BP \cap A \cap P') + n(BP \cap P \cap A') + n(BP \cap A \cap P) \\[3ex] = 5 + 17 + 17 + 20 \\[3ex] = 59 \\[3ex] $ 59 patients took (at least) blood pressure medication

(e.) Number of patients that took only pain medication = $n(P \cap A' \cap BP')$ = 26
26 patients took only pain medication.

(f.) Number of patients that exactly two medications = the sum of: Antibiotics and Blood pressure medication only and Antibiotics and Pain medication only and Blood pressure medication and Pain medication only

$ = n(A \cap BP \cap P') + n(A \cap P \cap BP') + n(BP \cap P \cap A') \\[3ex] = 17 + 22 + 17 \\[3ex] = 56 \\[3ex] $ 56 patients took exactly two medications.

(12.) Use the Venn diagram to answer questions (a.) through (d.)

(a.) How many people at the conference are unemployed women with a college degree?
(b.) How many people at the conference are employed men?
(c.) How many people at the conference are employed women without a college degree?
(d.) How many men are at the conference?

Let the set of:
women = W
college degree = D
currently employed = E

(a.) People at the conference that are unemployed women with a college degree = $n(W \cap D \cap E')$ women with college degree only = 19
19 people at the conference are unemployed women with a college degree

(b.) People at the conference that are employed men = $n(E \cap W' \cap D') + n(E \cap D \cap W')$ = the sum of: currently employed only and currently employed with a college degree only = 7 + 16 = 23
23 people at the conference are employed men

(c.) People at the conference are employed women without a college degree = $n(W \cap D \cap E')$ = currently employed women only = 8
8 people at the conference are employed women without a college degree

(d.) Men are at the conference = $n(W)'$ = the sum of: college degree only and currently employed only and currently employed with college degree only and neither women nor college degree nor currently employed = 9 + 7 + 16 + 3 = 35
35 men were at the conference

(14.) ACT A group of students was surveyed about what types of music each had listened to during the previous week.
Students could choose 1, 2, or 3 types of music (Rap, Rock, or Country), or they could respond "none of these music types."
Only 10 students responded with "none of these music types," and 18 students responded that they had listened to all 3 types.
In the figure below, the 8 regions show the numbers of responses in each category.

One region had exactly 8 responses.
What type(s) of music did those 8 students respond they had listened to during the previous week?
F. Rap only
G. Country and Rock only
H. Country and Rap only
J. Rap and Rock only
K. All 3 types of music

8 students had listened to Rap and Rock only ... Option J.

(15.) WASSCE A survey of 40 students showed that 23 students study Mathematics, 5 study Mathematics and Physics, 8 study Chemistry and Mathematics, 5 study Physics and Chemistry and 3 study all the three subjects.
The number of students who study Physics only is twice the number who study Chemistry only.

(a) Find the number of students who study:
(i) only Physics
(ii) only one subject

(b) What is the probability that a student selected at random studies exactly 2 subjects?

$ (a) \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] n(Physics\;\;ONLY) = k \\[3ex] n(Chemistry\;\;ONLY) = 2k \\[3ex] $

$ (i) \\[3ex] k + 2k + 13 + 2 + 2 + 5 + 3 = 40 \\[3ex] 3k + 25 = 40 \\[3ex] 3k = 40 - 25 \\[3ex] 3k = 15 \\[3ex] k = \dfrac{15}{3} \\[5ex] k = 5 \\[3ex] $ 5 students study only Physics

$ (ii) \\[3ex] n(only\;\;one\;\;subject) \\[3ex] = n(only\;\;Mathematics) + n(only\;\;Physics) + n(only\;\;Chemistry) \\[3ex] = 13 + k + 2k \\[3ex] = 13 + 5 + 2(5) \\[3ex] = 18 + 10 \\[3ex] = 28 \\[3ex] $ 28 students study only one subject

$ n(exactly\;\;two\;\;subjects) \\[3ex] = n(only\;\;Mathematics\;\;and\;\;Physics) + n(only\;\;Mathematics\;\;and\;\;Chemistry) + n(only\;\;Physics\;\;and\;\;Chemistry) \\[3ex] = 2 + 5 + 2 \\[3ex] = 9 \\[3ex] n(\xi) = 40 \\[3ex] P(exactly\;\;two\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;two\;\;subjects)}{n(\xi)} \\[5ex] = \dfrac{9}{40} $

(16.) JAMB

The shaded portion in the Venn diagram above is

$ A.\;\; X \cap Z \\[3ex] B.\;\; X^c \cap Y \cap Z \\[3ex] C.\;\; X \cap Y^c \cap Z \\[3ex] D.\;\; X \cap Y \cap Z^c \\[3ex] $

Explained in these resources:
Notes: Venn Diagrams for Three Sets: Examples 14 - 16

The shaded area is X and Z only
This implies: $X \cap Z \cap Y'$...Option C

(18.) JAMB

In the Venn diagram above, the shaded region is

$ A.\;\; (P \cap Q) \cup R \\[3ex] B.\;\; (P \cap Q) \cap R \\[3ex] C.\;\; (P \cap Q') \cap R \\[3ex] D.\;\; (P \cap Q') \cup R \\[3ex] $

Explained in these resources:
Notes: Venn Diagrams for Three Sets: Examples 14 - 16

The shaded area is P and R only
This implies:

$ P \cap R \cap Q' \\[3ex] = P \cap Q' \cap R ... Commutative\;\;Law \\[3ex] = (P \cap Q') \cap R ... Associative\;\;Law $

(19.) WASSCE In a road worthiness test on 240 cars, 60% passed.
The number that failed had faults in Clutch, Brakes, and Steering as follows: Clutch only - 28; Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6.
The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.

(a) Draw a Venn diagram to illustrate this information.

(b) How many cars had:
(i) faulty brakes?
(ii) only one fault?

$ n(\xi) = 240 \\[3ex] 60\% \;\;passed \implies 40\%\;\;failed \\[3ex] Passed:\;\; 60\% * 240 = 144 \\[3ex] Failed:\;\; 40\% * 240 = 96 \\[3ex] \underline{Failed} \\[3ex] Let: \\[3ex] Clutch = C \\[3ex] Brakes = B \\[3ex] Steering = T \\[3ex] n(B\;\;only) = p \\[3ex] \rightarrow n(T\;\;only) = 2p \\[3ex] $ (a.)
The Venn diagram is:

$ \underline{Failed} \\[3ex] p + 2p + 28 + 6 + 12 + 6 + 8 = 96 \\[3ex] 3p + 60 = 96 \\[3ex] 3p = 96 - 60 \\[3ex] 3p = 36 \\[3ex] p = \dfrac{36}{3} \\[5ex] p = 12 \\[3ex] (i) \\[3ex] n(B) = p + 12 + 6 + 8 \\[3ex] = 12 + 12 + 6 + 8 \\[3ex] = 38 \\[3ex] $ 38 cars had faulty brakes

$ (ii) \\[3ex] n(only\;\;one\;\;fault) \\[3ex] = n(only\;\;C) + n(only\;\;B) + n(only\;\;T) \\[3ex] = 28 + p + 2p \\[3ex] = 28 + 3p \\[3ex] = 28 + 3(12) \\[3ex] = 28 + 36 \\[3ex] = 64 \\[3ex] $ 64 cars had only one fault

(20.) GCSE Thirty shoppers were asked which of apples, oranges and bananas, they regularly liked to eat.
Of these 30 shoppers

2 liked all three fruits,
5 liked oranges but did not like apples or bananas,
6 liked both oranges and bananas,
17 liked apples,
10 liked bananas.

(a) Extra information from the 30 shoppers has been shown in the Venn diagram below.
Complete the Venn diagram using the information given above.

(b) One shopper is chosen at random.
Find the probability that this shopper does not like any of the three fruits.

(c) One shopper is chosen at random from those who like apples.
Find the probability that this shopper does not like oranges.

$ \underline{Given:\;\;in\;\;English} \\[3ex] n(\varepsilon) = 30 \\[3ex] n(all\;\;three\;\;fruits) = 2 \\[3ex] n(oranges\;\;but\;\;not\;\;apples\;\;or\;\;bananas) = 5 \\[3ex] n(oranges\;\;and\;\;bananas) = 6 \\[3ex] n(apples) = 17 \\[3ex] n(bananas) = 10 \\[3ex] \underline{Given:\;\;in\;\;Diagram} \\[3ex] n(apples\;\;and\;\;oranges\;\;but\;\;not\;\;bananas) = 8 \\[3ex] n(apples\;\;and\;\;bananas\;\;but\;\;not\;\;oranges) = 1 \\[3ex] \underline{Determine} \\[3ex] n(oranges\;\;and\;\;bananas\;\;but\;\;not\;\;apples) \\[3ex] = 6 - 2 \\[3ex] = 4 \\[3ex] n(apples\;\;but\;\;not\;\;oranges\;\;or\;\;bananas) \\[3ex] = 17 - (1 + 8 + 2) \\[3ex] = 17 - 11 \\[3ex] = 6 \\[3ex] n(bananas\;\;but\;\;not\;\;apples\;\;or\;\;oranges) \\[3ex] = 10 - (1 + 2 + 4) \\[3ex] = 10 - 7 \\[3ex] = 3 \\[3ex] n(neither\;\;apples\;\;nor\;\;oranges\;\;nor\;\;bananas) = p \\[3ex] 6 + 5 + 3 + 1 + 8 + 4 + 2 + p = 30 \\[3ex] 29 + p = 30 \\[3ex] p = 30 - 29 \\[3ex] p = 1 \\[3ex] $ (a) The Venn diagram is:

$ (b) \\[3ex] n(neither\;\;fruit) = p = 1 \\[3ex] n(\varepsilon) = 30 \\[3ex] P(neither\;\;fruit) = \dfrac{n(neither\;\;fruit)}{n(\varepsilon)} = \dfrac{1}{30} \\[5ex] (c) \\[3ex] \underline{From\;\;Apples} \\[3ex] n(apples) = 17 \\[3ex] n(apples\;\;but\;\;not\;\;oranges) = 6 + 1 = 7 \\[3ex] P(apples\;\;but\;\;not\;\;oranges) \\[3ex] = \dfrac{n(apples\;\;but\;\;not\;\;oranges)}{n(apples)} \\[5ex] = \dfrac{7}{17} $

Top

ACT Use the following information to answer questions 21 - 23
Each of the 200 people in a random sample of the 2,500 people at the mall today was asked which, if any, of the following types of pets he or she owns: bird, cat, dog, or fish.
All 200 people answered the question.
The answers were tallied, and the exact percents of people who own the pets are shown in the diagram below.

(21.) Because this was a random sample, the percents in the sample are the most likely estimates for the corresponding percents among all the people at the mall today.
What estimate does this give for the number of people at the mall today who own dogs but none of the other 3 types of pets?

$ F.\;\; 125 \\[3ex] G.\;\; 250 \\[3ex] H.\;\; 375 \\[3ex] J.\;\; 500 \\[3ex] K.\;\; 875 \\[3ex] $

Sample size = 200
Population size = 1200

Depening on time:
Discuss with your Statistics students regarding the conditions for a sample proportion to estimate population proportion.

Sample is a likely estimate of the population because the sample is a random sample
The number of people who own dogs but none of the other 3 types of pets means the number of people who own ONLY dogs

$ \underline{Population} \\[3ex] n(dogs\;\;ONLY) \\[3ex] 15\% * 1200 \\[3ex] = \dfrac{15}{100} * 1200 \\[5ex] = 375 \\[3ex] $ About 375 people at the mall today own only dogs.

(22.) What percent of the people in the random sample own exactly 1 type of the 4 types of pets?

$ A.\;\; 10\% \\[3ex] B.\;\; 25\% \\[3ex] C.\;\; 45\% \\[3ex] D.\;\; 70\% \\[3ex] E.\;\; 90\% \\[3ex] $

$ \underline{Sample} \\[3ex] n(exactly\;\;1\;\;type\;\;of\;\;the\;\;4\;\;types\;\;of\;\;pets) \\[3ex] = n(cat\;\;ONLY) \\[3ex] + n(dog\;\;ONLY) \\[3ex] + n(bird\;\;ONLY) \\[3ex] + n(fish\;\;ONLY) \\[3ex] = 5\% + 15\% + 5\% + 45\% \\[3ex] = 70\% \\[3ex] $ 70% of the 200 people in the random sample own excatly 1 type of the types of pet.

(23.) Suppose 25 additional people at random were asked the question, with the following answers: 15 own fish only, 5 own a cat and a dog only, and 5 own a cat, a dog, and a bird only.
Among all 225 people asked, what fraction own fish but none of the other 3 types of pets?

$ F.\;\; \dfrac{105}{225} \\[5ex] G.\;\; \dfrac{105}{215} \\[5ex] H.\;\; \dfrac{105}{200} \\[5ex] J.\;\; \dfrac{115}{225} \\[5ex] K.\;\; \dfrac{115}{200} \\[5ex] $

$ \underline{Initial\;\;Sample} \\[3ex] sample\;\;size = 200 \\[3ex] n(fish\;\;ONLY) \\[3ex] = 45\% * 200 \\[3ex] = \dfrac{45}{100} * 200 \\[5ex] = 90 \\[3ex] \underline{Additional\;\;Sample} \\[3ex] sample\;\;size = 25 \\[3ex] n(fish\;\;ONLY) = 15 \\[3ex] \underline{Total\;\;Sample} \\[3ex] sample\;\;size = 200 + 25 = 225 \\[3ex] n(fish\;\;ONLY) = 90 + 15 = 105 \\[3ex] Fraction(fish\;\;ONLY) = \dfrac{105}{225} $

(24.) CSEC The Venn Diagram below represents information on the type of games played by members of a youth club.
All members of the club play at least one game.

S represents the set of members who play squash.
T represents the set of members who play tennis.
H represents the set of members who play hockey.
Leo, Mia and Neil are three members of the youth club.

(i) State what game(s) is/are played by
(a) Leo
(b) Mia
(c) Neil

(ii) Describe in words the members of the set $H' \cap S$

(i)
(a) Leo plays tennis and hockey
He plays tennis
He plays hockey because of the intersection of T and H

(b) Mia plays squash, tennis, and hockey
She plays squash
She plays tennis because S is a subset of T
She plays hockey because of the intersection of S and H

(c) Neil plays only hockey

(ii)
The members of the set $H' \cap S$ are the members that play squash but not hockey
Because squash is a subset of tennis, all those members also play tennis.
In that regard, we can also describe them as the members that play squash and tennis but not hockey.

(25.) WASCCE In a class of 200 students, 70 offered Physics, 90 Chemistry, 100 Mathematics while 24 did not offer any of the three subjects.
Twenty three (23) students offered Physics and Chemistry, 41 Chemistry and Mathematics while 8 offered all three subjects.

(a) Draw a Venn diagram to illustrate the information.

(b) Find the probability that a student selected at random from the class offered:
(i) Physics only
(ii) Exactly two of the subjects

$ n(\mu) = 200 \\[3ex] Let: \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] Mathematics = M \\[3ex] n(P) = 70 \\[3ex] n(C) = 90 \\[3ex] n(M) = 100 \\[3ex] n(neither) = 24 \\[3ex] n(P\;\;and\;\;C) = 23 \\[3ex] n(C\;\;and\;\;M) = 41 \\[3ex] n(all\;\;three) = 8 \\[3ex] n(P\;\;and\;\;M\;\;only) = d \\[3ex] $ (a) The Venn diagram is:

$ n(P\;\;and\;\;C\;\;only) \\[3ex] = 23 - 8 \\[3ex] = 15 \\[3ex] n(C\;\;and\;\;M\;\;only) \\[3ex] = 41 - 8 \\[3ex] = 33 \\[3ex] n(C\;\;only) \\[3ex] = 90 - (15 + 8 + 33) \\[3ex] = 90 - 56 \\[3ex] = 34 \\[3ex] n(P\;\;only) \\[3ex] = 70 - (d + 8 + 15) \\[3ex] = 70 - (d + 23) \\[3ex] = 70 - d - 23 \\[3ex] = 47 - d \\[3ex] n(M\;\;only) \\[3ex] = 100 - (d + 8 + 33) \\[3ex] = 100 - (d + 41) \\[3ex] = 100 - d - 41 \\[3ex] = 59 - d \\[3ex] \implies \\[3ex] (47 - d) + (59 - d) + 34 + d + 15 + 33 + 8 + 24 = 200 \\[3ex] 47 - d + 59 - d + d + 34 + 15 + 33 + 8 + 24 = 200 \\[3ex] 220 - d = 200 \\[3ex] 220 - 200 = d \\[3ex] 20 = d \\[3ex] d = 20 \\[3ex] \implies \\[3ex] n(P\;\;and\;\;M\;\;only) = 20 \\[3ex] (b) \\[3ex] (i) \\[3ex] n(P\;\;only) = 47 - d = 47 - 20 = 27 \\[3ex] n(\mu) = 200 \\[3ex] P(P\;\;only) = \dfrac{n(P\;\;only)}{n(\mu)} = \dfrac{27}{200} \\[5ex] (ii) \\[3ex] n(exactly\;\;2\;\;subjects) \\[3ex] = n(P\;\;and\;\;C\;\;only) + n(C\;\;and\;\;M\;\;only) + n(P\;\;and\;\;M\;\;only) \\[3ex] = 15 + 33 + 20 \\[3ex] = 68 \\[3ex] P(exactly\;\;2\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;2\;\;subjects)}{n(\mu)} \\[5ex] = \dfrac{68}{200} \\[5ex] = \dfrac{17}{50} $

(26.) JAMB Which of the Venn diagrams below represents $P' \cap Q \cap R'$

Explained in these resources:
Notes: Venn Diagrams for Three Sets: Examples 14 - 16

$ P' \cap Q \cap R' = Q\;\;only \\[3ex] $ The shaded area for Q only is Option D.

(28.) WASCCE

In the Venn diagram, P, Q, and R are subsets of the universal set U
If $n(U) = 125$, find:
(a) the value of x
(b) $n(P \cup Q \cap R')$

$ (a) \\[3ex] (16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125 \\[3ex] 16 - 2x + 6 + x + 19 - 3x + 5x + 7x + 8x + 4x + 4 = 125 \\[3ex] 20x + 45 = 125 \\[3ex] 20x = 125 - 45 \\[3ex] 20x = 80 \\[3ex] x = \dfrac{80}{20} \\[5ex] x = 4 \\[3ex] (b) \\[3ex] $ We need to find the shaded area for $P \cup Q \cap R'$ before we can find the cardinality
Explained in these resources:
Notes: Venn Diagrams for Three Sets: Examples 14 - 16

Let us do this (draw a Venn diagram with some elements) quickly:

$ U = \{1, 2, 3, 4, 5, 6, 7, 8\} \\[3ex] P = \{1, 2, 4, 5\} \\[3ex] Q = \{2, 3, 5, 6\} \\[3ex] R = \{4, 5, 6, 7\} \\[3ex] P \cup Q = \{1, 2, 3, 4, 5, 6\} \\[3ex] R' = \{1, 2, 3, 8\} \\[3ex] (P \cup Q) \cap R' = \{1, 2, 3\} \\[3ex] $

$ \implies \\[3ex] n(P \cup Q \cap R') \\[3ex] = (16 - 2x) + 5x + (6 + x) \\[3ex] = 16 - 2x + 5x + 6 + x \\[3ex] = 4x + 22 \\[3ex] = 4(4) + 22 \\[3ex] = 16 + 22 \\[3ex] = 38 $

(30.) NSC A survey was conducted among 150 learners at a school.
The following observations were made:

The probability that a learner, selected at random, will take part in:
- Only hockey (H) is 0,24
- Hockey and debating (D), but not chess (C) is 0,14
- Debating and chess, but not hockey is 0,12
- Hockey and chess, but not debating is 0,02
The probability that a learner, selected at random, participates in at least one activity is 0,7
15 learners participated in all three activities.
The number of learners that participate only in debating is the same as the number of learners who participate only in chess.

The Venn diagram below shows some of the above information.

(30.1) Determine a, the probability that a learner, selected at random, participates in all three activities.

(30.2) Determine m, the probability that a learner, selected at random, does NOT participate in any of the three activities.

(30.3) How many learners play only chess?

$ (30.1) \\[3ex] a = \dfrac{n(H \cap D \cap C)}{n(S)} \\[5ex] = \dfrac{15}{150} = \dfrac{1}{10} = 0.1 \\[3ex] (30.2) \\[3ex] P(at\;\;least\;\;one\;\;activity) + P(no\;\;activity) = 1 \\[3ex] ...Complementary\;\;Rule \\[3ex] 0.7 + m = 1 \\[3ex] m = 1 - 0.7 \\[3ex] m = 0.3 \\[3ex] (30.3) \\[3ex] P(only\;\;1\;\;activity) = 0.24 + b + b = 2b + 0.24 \\[3ex] P(only\;\;2\;\;activities) = 0.02 + 0.14 + 0.12 = 0.28 \\[3ex] P(all\;\;three\;\;activities) = a = 0.1 \\[3ex] P(no\;\;activity) = m = 0.3 \\[3ex] P(total\;\;probabability) = 1 \\[3ex] \implies \\[3ex] 2b + 0.24 + 0.28 + 0.1 + 0.3 = 1 \\[3ex] 2b + 0.92 = 1 \\[3ex] 2b = 1 - 0.92 \\[3ex] 2b = 0.08 \\[3ex] b = \dfrac{0.08}{2} \\[5ex] b = 0.04 \\[3ex] P(only\;\;chess) = \dfrac{n(only\;\;chess)}{n(S)} \\[5ex] b = \dfrac{n(only\;\;chess)}{150} \\[5ex] n(only\;\;chess) \\[3ex] = 150 * b \\[3ex] = 150 * 0.04 \\[3ex] = 6 \\[3ex] $ 6 learners play only chess

(31.) In the given Venn Diagram, sets are labeled with uppercase letters, A, B, C and U, and regions are labeled with lowercase letters p, q, r, s, t, u, and v.
Identify the region(s) that result from performing the given set operations.

(a.) List the region(s) included in A'
(b.) List the region(s) included in B ⋃ C
(c.) Use the answers in parts (a.) and (b.) to decide which of the shaded region(s) represents the set A' ⋂ (B ⋃ C)

There are at least two approaches that we can use to solve this question.
We can do a visual inspection or we can list the elements on paper (as we did in the Notes)
Let us use the easiest approach: By Inspection
But you can list them if you prefer.

(a.) A' includes all the elements not in set A
A' = {p, q, r, w}

(b.) B ⋃ C includes all the elements in set B and in set C
B ⋃ C = {p, u, t, q, s, r}

(c.) A' ⋂ (B ⋃ C) includes the common elements in A' and (B ⋃ C)
This means the elements that are common to parts (a.) and (b.)
A' ⋂ (B ⋃ C) = {p, q, r}
The correct answer is Option D.

(33.) WASCCE In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only.
If each student offered at least one of the subjects,

(a) how many students failed in all subjects?

(b) find the percentage number that failed in at least one of Economics and Mathematics

(c) calculate the probability that a student picked at random failed in Accounts.

$ n(\mu) = 40 \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Accounts = A \\[3ex] Economics = E \\[3ex] n(M) = 18 \\[3ex] n(A) = 19 \\[3ex] n(E) = 16 \\[3ex] n(M\;\;and\;\;A\;\;only) = 5 \\[3ex] n(M\;\;only) = 6 \\[3ex] n(A\;\;only) = 9 \\[3ex] n(A\;\;and\;\;E\;\;only) = 2 \\[3ex] n(M\;\;and\;\;A\;\;and\;\;E) = n(passed\;\;all\;\;three) \\[3ex] $ Let us use a Venn diagram

$ \underline{Accounts} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(A\;\;and\;\;E\;\;only) + n(A\;\;only) = n(A) \\[3ex] n(passed\;\;all\;\;three) + 5 + 2 + 9 = 19 \\[3ex] n(passed\;\;all\;\;three) + 16 = 19 \\[3ex] n(passed\;\;all\;\;three) = 19 - 16 \\[3ex] n(passed\;\;all\;\;three) = 3 \\[3ex] \underline{Mathematics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(M\;\;and\;\;E\;\;only) + n(M\;\;only) = n(M) \\[3ex] 3 + 5 + n(M\;\;and\;\;E\;\;only) + 6 = 18 \\[3ex] 14 + n(M\;\;and\;\;E\;\;only) = 18 n(M\;\;and\;\;E\;\;only) = 18 - 14 \\[3ex] n(M\;\;and\;\;E\;\;only) = 4 \\[3ex] \underline{Economics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) + n(E\;\;only) = n(E) \\[3ex] 3 + 4 + 2 + n(E\;\;only) = 16 \\[3ex] 9 + n(E\;\;only) = 16 \\[3ex] n(E\;\;only) = 16 - 9 \\[3ex] n(E\;\;only) = 7 \\[3ex] \implies \\[3ex] (a) \\[3ex] n(M\;\;only) + n(A\;\;only) + n(E\;\;only) \\[3ex] + n(M\;\;and\;\;A\;\;only) + n(N\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) \\[3ex] + n(passed\;\;all\;\;three) + n(failed\;\;all\;\;three) = n(\mu) \\[3ex] 6 + 9 + 7 + 5 + 4 + 2 + 3 + n(failed\;\;all\;\;three) = 40 \\[3ex] 36 + n(failed\;\;all\;\;three) = 40 \\[3ex] n(failed\;\;all\;\;three) = 40 - 36 \\[3ex] n(failed\;\;all\;\;three) = 4 \\[3ex] $ 4 students failed all subjects.

The completed Venn diagram is:

(b) Failed at least one of Economics and Mathematics implies:
failed in Economics only
failed in Mathematics only
failed in Economics and Mathematics only

failed in all three subjects

In other words, it implies: not passing anything that has to do with Economics or Mathematics
Because of the Venn diagram (that shows passing for the subjects):
This means:
passing only Accounts
failing all three subjects

$ n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = n(A\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 9 + 4 \\[3ex] = 13 \\[3ex] n(U) = 40 \\[3ex] \%(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = \dfrac{n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics)}{n(U)} * 100 \\[5ex] = \dfrac{13}{40} * 100 \\[5ex] = 32.5\% \\[3ex] (c) \\[3ex] P(failed\;\;in\;\;Accounts) + P(passed\;\;in\;\;Accounts) = 1 ...Complementary\;\;Rule \\[3ex] P(failed\;\;in\;\;Accounts) = 1 - P(passed\;\;in\;\;Accounts) \\[3ex] P(passed\;\;in\;\;Accounts) = \dfrac{n(A)}{n(U)} = \dfrac{19}{40} \\[5ex] \implies \\[3ex] P(passed\;\;in\;\;Accounts) = 1 - \dfrac{19}{40} = \dfrac{40 - 19}{40} = \dfrac{21}{40} \\[5ex] $ Alternatively:
Failed in Accounts means that the student did not pass anything related to Accounts
It is $n(A')$
Because of the Venn diagram (that shows passing for the subjects):
This includes:
passing only Mathematics
passing only Economics
passed Mathematics and Economics only
failing all three subjects

$ n(failed\;\;in\;\;Accounts) \\[3ex] = n(M\;\;only) + n(E\;\;only) + n(M\;\;and\;\;E\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 6 + 7 + 4 + 4 \\[3ex] = 21 \\[3ex] n(U) = 40 \\[3ex] P(failed\;\;in\;\;Accounts) \\[3ex] = \dfrac{n(failed\;\;in\;\;Accounts)}{n(U)} \\[5ex] = \dfrac{21}{40} $

(34.) JAMB P, Q and R are subsets of the universal set, U
The Venn diagram showing the relationship $(P \cap Q) \cup R$ is

Explained in these resources:
Notes: Venn Diagrams for Three Sets: Examples 14 - 16

Let us do this (draw a Venn diagram with some elements) quickly:

$ P = \{1, 2, 4, 5\} \\[3ex] Q = \{2, 3, 5, 6\} \\[3ex] R = \{4, 5, 6, 7\} \\[3ex] P \cap Q = \{2, 5\} \\[3ex] (P \cap Q) \cup R = \{2, 4, 5, 6, 7\} \\[3ex] $

The correct answer is Option C

(39.) A media company that produces videos surveyed its customers to determine whether they would prefer to purchase the videos by buying an access code to allow them to download from a website, purchase hard copies on a DVD, or access the videos via a podcast.
The company has extracted the table of information shown below from its customer database:

	Access Code (A)	Podcast (P)	DVD (D)	Total
Under 41 (Y)oung	73	50	48	171
41 – 55 (M)iddle aged	35	43	55	133
Over 55 (S)enior	33	70	55	158
Total	141	163	158	462

Use the information in the table to find the number of elements in:
(a.) P
(b.) D ⋂ M
(b.) A ⋃ M
(c.) S'
(e.) M ⋂ (A ⋃ D)

$ (a.) \\[3ex] n(P) = 163 \\[3ex] (b.) \\[3ex] n(D \cap M) = 55 \\[3ex] (c.) \\[3ex] n(A \cup M) \\[3ex] = n(A) + n(M) - n(A \cap M)...Addition\;\;Rule \\[3ex] = 141 + 133 - 35 \\[3ex] = 239 \\[3ex] (d.) \\[3ex] n(S') \\[3ex] = n(Y) + n(M) \\[3ex] = 171 + 133 \\[3ex] = 304 \\[3ex] (e.) \\[3ex] n[M \cap (A \cup D)] \\[3ex] = n[(M \cap A) \cup (M \cap D)]...Distributive\;\;Law \\[3ex] = 35 + 55 \\[3ex] = 90 \\[3ex] $

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