 # Union and Intersection of Sets of Cardinal Numbers

Cardinal Number of a Set

The cardinal number of a finite set is the number of distinct elements within the set. In other words, the cardinal number of a set represents the size of a set.

The cardinal number of a set named M, is denoted as n(M). Here, M is the set and n(M) is the number of elements in set M.

### a union b

A union of sets is when two or more sets are taken together and grouped. If set M and set N are a union, then it is written as M ∪ N.

Disjoint Sets: Disjoint sets are sets that have no elements in common and do not intersect. If M and N are disjoint sets, then it can be mathematically represented as M ∩ N = ∅.

a union b formula

If M and N are finite sets and they are disjoint, then the sum of the cardinal numbers of M and N will be the cardinal number of the union of sets M and N.

n(M ∪ N) = n(M) + n(N)

a intersection b

Intersection of Sets: Two sets intersect when they have one or more common elements. Each coon element is a point of intersection for the two sets.

a intersection b formula

When two sets (M and N) intersect, then the cardinal number of their union can be calculated in two ways:

1.  The cardinal number of their union is the sum of their cardinal numbers of the individual sets minus the number of common elements.

n(M ∪ N) = n(M) + n(N) - n(M ∩ N)

2.  The cardinal number of their union is given by the sum of their uncommon  elements and their common elements.

n (M ∪ N) = n (M – N) + n(N – M) + n(M ∩ N)

### Union and Intersection of Three Sets

If M, N, and C are three finite sets that intersect each-other and are in union, their cardinal number can be represented as n(M ∪ N ∪ C).

Union and Intersection of Three Sets Formula

The cardinal number of the union of three sets is the sum of the cardinal numbers of each individual set and the common elements of all three sets, excluding the common elements of pairs of sets.

n(M ∪ N ∪ C) = n(M) + n(N) + n(C) – n(M ∩ N) – n(N ∩ C) – n(M ∩ C) + n(M ∩ N ∩ C)

Probability Union and Intersection

Probability of Union

The probability for a union of sets depends on the compatibility of the events.

Sum Rule: Two events are said to be incompatible events if they are mutually exclusive and cannot occur simultaneously. The probability of incompatible events is given by the sum of the probabilities of the two events.

P(M ∪ N) = P(M) + P(N)

Compatible events are those events that may occur together and are not mutually exclusive. Their probability can be calculated by taking the sum of probabilities of the events and subtracting the times where they occur together.

P(O ∪ G) = P(O) + P(G) - P(O ∩ G)

Probability of Intersection

Product Rule: If two events M and event N must happen in order for a certain outcome to occur, and if M and N are independent events, then the probabilities can be calculated by multiplying the probabilities of M and N.

P(M ∩ N)=P(M)*P(N)

The probability of two dependent events occurring together is given by: P(M ∩ N)=P(M/N)*P(N)

### Venn Diagram Union and Intersection Problem Example

Example: There are a total of 200 boys in class XII. 120 of them study math, 50 students study science and 30 students study both mathematics and science. Find the number of boys who

n(Total) = 200

n(Math) = 120

n(Science) = 50

n(Math ∩ Science) = 30

(i)Study math but not science

Students who study math but not science is basically the total number of math students minus the number of students who study both science and math.

n(only Math)= n(Math)-n(Both)

n(only Math)=120-30

n(only Math)=90

(ii)Study science but not math

Students who study science but not math is basically the total number of science students minus the number of students who study both science and math.

n(only Science)= n(Science)-n(Both)

n(only Science)=50-30

n(only Science)=20

(iii)Study math or science

Students who study science or math is basically the total number of science students plus the total number of math students minus the students who study both science and math.

n(Science/Math)= n(Science)+n(Math)-n(Both)

n(Science/Math)=50+120-30

n(Science/Math)=140