Laws of Set Theory in Probability

These laws of set theory are foundational principles used to understand relationships between sets and are directly applicable in probability theory.

Commutative Laws

The commutative laws state that the order of union and intersection of sets does not matter:

• Union: A∪B = B∪A
• Intersection: A∩B = B∩A

Example: Let A={1,2,3} and B={3,4,5}. Show that A∪B = B∪A and A∩B = B∩A.

A∪B = {1,2,3}∪{3,4,5} = {1,2,3,4,5}
B∪A = {3,4,5}∪{1,2,3} = {1,2,3,4,5}.

A∩B = {1,2,3}∩{3,4,5} = {3}
B∩A = {3,4,5}∩{1,2,3} = {3}.

Associative Laws

The associative laws state that the way sets are grouped in union and intersection does not affect the result:

• Union: (A∪B)∪C = A∪(B∪C)
• Intersection: (A∩B)∩C = A∩(B∩C)

Example: Let A={1,2}, B={2,3}, and C={3,4}Solution:

A∩(B∩C) = {1,2}∩({2,3}∩{3,4}) = {1,2}∩{3} = ∅ (empty set).

(A∪B)∪C = ({1,2}∪{2,3})∪{3,4} = {1,2,3}∪{3,4} = {1,2,3,4}
A∪(B∪C)={1,2}∪({2,3}∪{3,4})={1,2}∪{2,3,4}={1,2,3,4}. Thus, (A∪B)∪C=A∪(B∪C).

(A∩B)∩C = ({1,2}∩{2,3})∩{3,4} = {2}∩{3,4} = ∅
A∩(B∩C) = {1,2}∩({2,3}∩{3,4}) = {1,2}∩{3} = ∅. Thus, (A∩B)∩C=A∩(B∩C).

Distributive Laws

The distributive laws describe how union and intersection distribute over each other:

• Union: A∪(B∩C)=(A∪B)∩(A∪C)
• Intersection: A∩(B∪C)=(A∩B)∪(A∩C)

Example: Let A={1,2}, B={2,3}, and C={3,4}. Show that A∪(B∩C)=(A∪B)∩(A∪C) and A∩(B∪C)=(A∩B)∪(A∩C).

B∩C = {2,3}∩{3,4} = {3}
A∪(B∩C) = {1,2}∪{3} = {1,2,3}
A∪B = {1,2}∪{2,3} = {1,2,3}
A∪C = {1,2}∪{3,4} = {1,2,3,4}
(A∪B)∩(A∪C)={1,2,3}∩{1,2,3,4}={1,2,3}. Thus, A∪(B∩C)=(A∪B)∩(A∪C).

B∪C = {2,3}∪{3,4} = {2,3,4}
A∩(B∪C) = {1,2}∩{2,3,4} = {2}
A∩B = {1,2}∩{2,3} = {2}
A∩C = {1,2}∩{3,4} =∅
(A∩B)∪(A∩C) = {2}∪∅ = {2}. Thus, A∩(B∪C) = (A∩B)∪(A∩C).

De Morgan’s Laws

De Morgan’s laws describe the complement of the union and intersection of two sets:

• Union Complement: (A∪B)^c = A^c∩B^c
• Intersection Complement: (A∩B)^c = A^c∪B^c

Example: Let U = {1,2,3,4,5,6} be the universal set, A = {1,2}, and B = {2,3}. Show that (A∪B)^c = A^c∩B^c and (A∩B)^c = A^c∪B^c.

A∪B = {1,2}∪{2,3} = {1,2,3}
A^c = U−A = {1,2,3,4,5,6}−{1,2} = {3,4,5,6}
B^c = U−B = {1,2,3,4,5,6}−{2,3} = {1,4,5,6}
(A∪B)^c = U−(A∪B) = {1,2,3,4,5,6}−{1,2,3} = {4,5,6}
A^c∩B^c = {3,4,5,6}∩{1,4,5,6} = {4,5,6}. Thus, (A∪B)^c=A^c∩B^c.

A∩B = {1,2}∩{2,3} = {2}
(A∩B)^c = U−(A∩B) = {1,2,3,4,5,6}−{2} = {1,3,4,5,6}.
A^c = U−A = {1,2,3,4,5,6}−{1,2} = {3,4,5,6}.
B^c = U−B = {1,2,3,4,5,6}−{2,3} ={ 1,4,5,6}.
A^c∪B^c = {3,4,5,6}∪{1,4,5,6} = {1,3,4,5,6}. Thus, (A∩B)^c=A^c∪B^c={1,3,4,5,6}.