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Are Complements of Independent Events Also Independent?

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Last Updated on December 13, 2025 by Rajeev Bagra

In probability theory, independence is a fundamental concept. A natural question that follows is:

If two events (A) and (B) are independent, are their complements also independent?

The answer is yes — always, for two events.
This post explains why and illustrates the result with clear examples.

Definition of Independence

Two events (A) and (B) are said to be independent if:

P(A \cap B)=P(A),P(B)

This means the occurrence of one event does not affect the probability of the other.

Complements and Independence

Let (A^c) and (B^c) denote the complements of (A) and (B).

Using basic probability rules:

P(A^c)=1-P(A), \quad P(B^c)=1-P(B)

If (A) and (B) are independent, then:

P(A^c \cap B^c)=(1-P(A))(1-P(B))=P(A^c),P(B^c)

Hence, the complements of independent events are also independent.

Example 1: Coin Toss and Die Roll

Consider two independent experiments.

  • Event (A): A fair coin shows Heads
  • Event (B): A fair die shows an even number
P(A)=\tfrac12,\quad P(B)=\tfrac12 P(A\cap B)=\tfrac14=\tfrac12\times\tfrac12

Now consider the complements:

  • (A^c): Coin shows Tails
  • (B^c): Die shows an odd number
P(A^c)=\tfrac12,\quad P(B^c)=\tfrac12 P(A^c\cap B^c)=\tfrac14=\tfrac12\times\tfrac12

Thus, the complements are independent.

Example 2: Two Independent Coin Tosses

  • Event (A): First coin is Heads
  • Event (B): Second coin is Heads
P(A)=P(B)=\tfrac12 P(A\cap B)=\tfrac14

Complements:

  • (A^c): First coin is Tails
  • (B^c): Second coin is Tails
P(A^c\cap B^c)=\tfrac14=P(A^c),P(B^c)

Again, independence is preserved.

Example 3: Business System Reliability

Suppose two independent systems operate in a company.

  • Event (A): Payment gateway succeeds
  • Event (B): Email confirmation is delivered
P(A)=0.99,\quad P(B)=0.98 P(A\cap B)=0.99\times0.98

Complements:

  • (A^c): Payment fails
  • (B^c): Email fails
P(A^c)=0.01,\quad P(B^c)=0.02 P(A^c\cap B^c)=0.01\times0.02=P(A^c),P(B^c)

Thus, failures are also independent.

Example 4: Marketing Campaign Clicks

  • Event (A): User clicks an email campaign
  • Event (B): User clicks a display ad

Assuming independence:

P(A)=0.3,\quad P(B)=0.2 P(A\cap B)=0.06

Complements:

  • (A^c): No email click
  • (B^c): No ad click
P(A^c)=0.7,\quad P(B^c)=0.8 P(A^c\cap B^c)=0.56=P(A^c),P(B^c)

What About Mixed Complements?

Independence also holds for mixed cases:

P(A\cap B^c)=P(A),P(B^c) P(A^c\cap B)=P(A^c),P(B)

So all four combinations remain independent.

Final Takeaway

If two events (A) and (B) are independent, then:

  • (A) and (B)
  • (A^c) and (B)
  • (A) and (B^c)
  • (A^c) and (B^c)

are all independent.

This result is always true for two events, and it is a useful property in probability modeling, statistics, business analytics, and system reliability analysis.

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