In this post we will discuss important formulas of sets which will help you solve questions of set theory and probability. Apart from the formulas, i have also solved some questions which will help to further increase your understanding about the chapter of set theory.
Also remember that regarding set theory, the concept of Venn diagram is also helpful to solve problems faster, so sure to use Venn diagram to solve as many questions as possible.
Formula of Union of sets
Formula of Union when two sets are distinct
Let A & B are two sets with no common element
A = { 1, 2, 3, 4}
B = { 5, 6, 7, 8}
Hence A ∩ B = φ,
Then Union of A & B is equal to the sum of element of set A and set B
n ( A ∪ B ) = n ( A ) + n ( B ) —- eq (1)
This is the formula for union of sets if the sets are distinct
Considering above example of set A & B, we will use the above formula
( A ∪ B ) = {1, 2, 3, 4, 5, 6, 7, 8)
Formula of union when two sets are not distinct
Now what if the sets are not distinct.
Consider the below sets
A = { 1, 2, 3, 4,}
B = { 4, 5, 6, 7}
Here you can see that the sets A & B are not distinct as they have common element 4 in between them.
Then the formula for ( A ∪ B ) is:
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
Considering above example of A & B
n ( A ∪ B ) = { 1, 2, 3, 4} + {4, 5, 6, 7} – {4}
n ( A ∪ B ) = {1, 2, 3, 4, 5, 6, 7)
Formula of union when there are three non-distinct sets
Let there are three sets A, B and C
A= {1, 2, 3, 4}
B = {4, 5, 6, 7}
C = {7, 8, 9, 10}
Then the formula for union will be
n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )
Now lets get the union of above sets A, B and C using the above formula
( A ∪ B ∪ C ) = { 1, 2, 3, 4} + {4, 5, 6, 7} + {7, 8, 9, 10} – {4} – {7} – {0} + 0
( A ∪ B ∪ C ) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
All the three formulas mentioned above are extremely important as they would help us to solve questions of sets and probability. So make sure you invest some time to remember them
Set Questions fully solved
(01) If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have ?
Given
X ∪ Y = 50
n (X) = 28
n (Y) = 32
To find
X ∩ Y
Solution
Let us recount the formula of union of two sets in which the sets are not distinct
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
Putting the values in the formula
50 = 28 + 32 – n ( X ∩ Y )
n ( X ∩ Y ) = 60 – 50
n ( X ∩ Y ) = 10
Hence ( X ∩ Y ) has 10 elements
(02) In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics ?
Given
n (math) = 12
n (math ∪ physics) = 20
n (math ∩ physics) = 4
To find:
Number of teacher who teach physics
Solution
Using the formula of union of two sets
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
20 = 12 + n (physics) – 4
n (physics) = 20 – 8
n (physics) = 12
Hence there are 12 teachers that can teach physics
(03) In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Given
n (U) =400
n (A) = 100
n (B) = 150
n (A ∩ B) = 75
Using the union formula
n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )
n ( A ∪ B ) = 100 + 150 – 75
n ( A ∪ B ) = 175
There are 175 people who drink either apple or orange juice
We have to find people who doesn’t drink either apple or orange n (A′ ∩ B′)
n (A′ ∩ B′) = n (A ∪ B)′
= n (U) – n (A ∪ B)
= n (U) – 175
= 400 – 175 => 225
Hence there are 225 people who does not drink with apple or orange juice
(04) In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Given
n (C ∪ T) = 65
n (cricket) = 40
n (C ∩ T) = 10
Using the union formula
n ( C ∪ T ) = n ( C ) + n ( T ) – n ( C ∩ T )
65= 40 + n(T) -10
65 = 30 + n(T)
n(T) = 35 people
There are 35 people who like Tennis
But we have to find number of people who like tennis only
n (tennis only) = n(T) – n (C ∩ T)
n (tennis only) = 35 – 10 = 25
Thus there are 25 people who like only tennis
Check the following Venn Diagram for better understanding
