Lesson Objectives

Dear learner,

At the end of the lesson, you will be able to:

• Apply rules and principles of set theory for practical situations.
• Use sets on real life applications

Application

Application of Sets: Sets are used to organize, group, and analyze data in various fields such as mathematics, computer science, and statistics. They help solve problems related to classification, sorting, probability, logic, and database operations by defining relationships between distinct objects.

Key Terms

• Intersection of sets
• Union of sets
• Complement of sets

Application Questions

1. In a study of 250 people, it was found that 150 people consume injera daily, 120 people consume bread daily, and 80 consume both injera and bread daily. How many people consume injera or bread but not both?
2. In a survey of 200 people in Addis Ababa, 120 prefer using buses for transportation, 100 prefer taxis, and 60 use both buses and taxis. How many people prefer either buses or taxis but not both?

Solution

1. Given:- Total number of people surveyed: $(n = 250)$

Number of people who consume injera daily: $(n(A) = 150)$

Number of people who consume both injera and bread daily: $(n(A \cap B) = 80)$

We need to find the number of people who consume either injera or bread but not both.

Step 1: Calculate the total number of people who consume either injera or bread (or both).

$n(A \cup B) = n(A) + (B) – n(A \cap B)$

$n(A \cup B) = 150 + 120 – 80 = 190$.

So, 190 people consume either injera or bread (or both).

Step 2: Calculate the number of people who consume either injera or bread but not both.

The number of people who consume only injera:

$n(A) – (A \cap B) = 150 – 80 = 70$

The number of people who consume only bread:

$n(B) – n(A \cap B) = 120 – 80 = 40$

Step 3: Add the people who consume only injera and those who consume only bread.

People who consume either injera or bread but not both $= 70 + 40 = 110$

Final Answer: The number of people who consume either injera or bread but not both is 110.

2. Given:

• Total number of people surveyed: $n = 200$
• Number of people who prefer buses: $n(A) = 120$
• Number of people who prefer taxis: $n(B) = 100$
• Number of people who prefer both buses and taxis: $n(A \cap B) = 60$

We need to find the number of people who prefer either buses or taxis but not both

Step 1: Find the number of people who prefer either buses or taxis (or both).

$n(A \cup B) = n(A) + n(B) – n(A \cap B)$
$n(A \cup B) = 120 + 100 – 60 = 160$

So, 160 people prefer either buses or taxis (or both).

Step 2: Find the number of people who prefer either buses or taxis but not both.

The number of people who prefer buses only:

$n(A) – n(A \cap B) = 120 – 60 = 60$

The number of people who prefer taxis only:

$n(B) – n(A \cap B) = 100 – 60 = 40$

Step 3: Add the people who prefer buses only and those who prefer taxis only.

{People who prefer either buses or taxis but not both} $= 60 + 40 = 100$.

The number of people who prefer either buses or taxis but not both is $100$.

Note:

For any two finite sets 𝐴 and 𝐵,

i. $n(A \cup B) = n(A) + n(B) − n(A \cap B)$,  

ii. $n(A / B) = n(A) – n (A \cap B)$.

iii. $n(A \Delta B) = n(A) + n(B) – 2n(A \cap B)$

Example:

1. In a class of 31 students, 22 students study physics, 20 students study chemistry and 5 students study neither. Calculate the number of students who study both subjects

2. Suppose A and B are sets such that $A \cup B$ has 20 elements, $A \cap B$ has 7 elements, and the number of elements in B is twice that of A. What is the number of elements in: a) A? b) B?.

3. Of 100 staff members of a school, 48 drink coffee, 25 drink both tea and coffee and everyone drinks either coffee or tea. How many staff members drink tea?

Solution

1. Given that $n(U) = 31, n(P) = 22, n(C) = 20, n(P \cup C)’ = U – (P \cup C) = 5$ Required: $n(P \cap C) = ?$

$n(P \cup C) = U – n(P \cup C)’ = 31 – 5 = 26$, $n(P \cup C) = n(P) + n(C) – n(P \cap C)$, $26 = 22 + 20 – n(P \cap C)$, Thus, $n(P \cap C) = 16$,

Therefore there are 16 students who study both physics and chemistry.

2. Given that $n (A \cup B) = 20$, $n (A \cap B) = 7$, $n (B) = 2 n(A)$ Required: a) $A?$ b) $B?$

$n (A \cup B) = n(A) + n (B) – n (A \cap B)$, $n (A \cup B) = n(A) + 2 n(A) – n (A \cap B)$, $n (A \cup B) = 3n(A) – n (A \cap B)$, $20 = 3n(A) – 7$,

$3n(A) = 27$, $n (A) = 9$, on the other hand $n (B) = 2 n(A) = 2 (9) = 18$.

Thus $n(B) = 18$.

3. Given that $n(U) = 100$, $n(C) = 48$, $n(C \cap T) = 25$, $n(P \cup C) = U = 100$.

Required: $n(T) = ?$

$n(C \cup T) = n(C) + n(T) – n(C \cap T)$, $100 = 48 + n(T) – 25$, $n( T ) = 100 – 48 + 25 = 77$.

Therefore $77$ staff members who drink tea.