Independence of Complements

Why Complements Matter

An event and its complement describe two exhaustive, mutually exclusive possibilities: either the event occurs or it does not. When two events are independent, it is natural to ask whether replacing one or both of them by their complements changes that relationship. It does not. This fact lets us calculate probabilities involving “not” events without rebuilding the probability model from scratch.

For example, suppose two independently occurring review flags are recorded for a hypothetical portfolio application. If one flag is “missing documentation,” its complement is “documentation is not missing.” The complement result allows us to calculate probabilities involving either flag or its absence directly from the two original event probabilities.

Definition: For events $A$ and $B$, the complement $A^c$ consists of all outcomes in the sample space that are not in $A$. The events $A$ and $B$ are independent when $$ P(A\cap B)=P(A)P(B). $$ The same definition applies to any other pair of events, including $A^c$ and $B$.

The defining equation is a statement about a joint probability: the probability that both events occur equals the product of their individual probabilities. Independence is not the same as mutual exclusivity. The complement result below concerns what happens when we replace an event in an already independent pair by its complement.

The Complement Independence Result

If $A$ and $B$ are independent, then each of the following pairs is independent: $A^c$ and $B$, $A$ and $B^c$, and $A^c$ and $B^c$. The result follows from the independence equation together with the fact that an event and its complement partition the sample space.

Theorem: If $A$ and $B$ are independent, then $$ P(A^c\cap B)=P(A^c)P(B), \qquad P(A\cap B^c)=P(A)P(B^c), $$ and $$ P(A^c\cap B^c)=P(A^c)P(B^c). $$ Thus any pair formed by choosing either an event or its complement from each of two independent events is also independent.

Here is a derivation for all three statements. Since $A\cap B$ and $A^c\cap B$ are disjoint and together make up $B$, their probabilities add to $P(B)$. Independence gives $P(A\cap B)=P(A)P(B)$. Subtracting that joint probability from $P(B)$ gives the first identity:

\begin{aligned} P(A^c\cap B) &=P(B)-P(A\cap B)\\ &=P(B)-P(A)P(B)\\ &=(1-P(A))P(B)\\ &=P(A^c)P(B). \end{aligned}

The same partition argument, now partitioning $A$ according to whether $B$ occurs, gives the second identity:

\begin{aligned} P(A\cap B^c) &=P(A)-P(A\cap B)\\ &=P(A)-P(A)P(B)\\ &=P(A)(1-P(B))\\ &=P(A)P(B^c). \end{aligned}

Finally, $A^c$ can be partitioned into $A^c\cap B$ and $A^c\cap B^c$. Subtract the first joint probability from $P(A^c)$, then use the first identity:

\begin{aligned} P(A^c\cap B^c) &=P(A^c)-P(A^c\cap B)\\ &=P(A^c)-P(A^c)P(B)\\ &=P(A^c)(1-P(B))\\ &=P(A^c)P(B^c). \end{aligned}

Each derived joint probability factors into the corresponding marginal probabilities. That is exactly the definition of independence for the relevant pair. The argument uses ordinary event probability rules and remains valid when an event has probability zero or one.

Calculating With Complemented Events

A useful way to organize calculations is to write down the probability of each event and its complement before finding any joint probabilities. If $A$ and $B$ are independent, the four possible combinations have probabilities obtained by multiplying the relevant marginal probabilities:

Combination	Probability
$A\cap B$	$P(A)P(B)$
$A\cap B^c$	$P(A)P(B^c)$
$A^c\cap B$	$P(A^c)P(B)$
$A^c\cap B^c$	$P(A^c)P(B^c)$

These four combinations partition the sample space: exactly one of them occurs on any outcome. Consequently, their probabilities must sum to one. This provides a useful arithmetic check, but it does not replace checking the independence assumption.

Worked Example: Two Independent Application Review Flags

In a hypothetical application review, let $A$ be the event that an application has a documentation flag, and let $B$ be the event that it has a coding flag. Assume the flags are independent, with $P(A)=\frac{1}{5}$ and $P(B)=\frac{3}{10}$. Find the probability that neither flag occurs, the probability that the coding flag occurs without the documentation flag, and $P(A\mid B^c)$.

Method. First calculate the complement probabilities, then use complement independence for the joint probabilities. For the conditional probability, divide the relevant joint probability by the probability of the conditioning event.

The complement probabilities are $$ P(A^c)=1-\frac{1}{5}=\frac{4}{5}, \qquad P(B^c)=1-\frac{3}{10}=\frac{7}{10}. $$

Neither flag occurs when $A^c$ and $B^c$ both occur. Since the original events are independent, their complements are independent: $$ P(A^c\cap B^c)=\frac{4}{5}\cdot\frac{7}{10} =\frac{28}{50}=\frac{14}{25}=0.56. $$

The coding flag occurs without the documentation flag when $A^c\cap B$ occurs: $$ P(A^c\cap B)=\frac{4}{5}\cdot\frac{3}{10} =\frac{12}{50}=\frac{6}{25}=0.24. $$

For the conditional probability, the denominator is $P(B^c)=\frac{7}{10}$, which is positive. The numerator is $P(A\cap B^c)=P(A)P(B^c)$: $$ P(A\mid B^c) =\frac{P(A\cap B^c)}{P(B^c)} =\frac{\left(\frac{1}{5}\right)\left(\frac{7}{10}\right)}{\frac{7}{10}} =\frac{1}{5}=0.20. $$

Check. The four joint probabilities are $\frac{3}{50}$, $\frac{7}{50}$, $\frac{12}{50}$, and $\frac{28}{50}$. Their sum is $\frac{50}{50}=1$. The conditional result also matches $P(A)$, consistent with $A$ being independent of $B^c$. Common mistake avoided: using $P(A)P(B)$ for the “neither flag” probability; that product applies to $A\cap B$, not $A^c\cap B^c$.

A Complement Does Not Mean “Subtract the Intersection”

A frequent error is to calculate $P(A^c\cap B^c)$ as $1-P(A\cap B)$. The complement of $A\cap B$ is the event that at least one of $A$ and $B$ does not occur; it includes outcomes in which one event occurs and the other does not. It is therefore generally larger than $A^c\cap B^c$, which requires both events not to occur.

Common trap: The complement of $A\cap B$ is $(A\cap B)^c$, not $A^c\cap B^c$. In fact, $(A\cap B)^c=A^c\cup B^c$. For independent $A$ and $B$, calculate the probability that neither occurs as $P(A^c)P(B^c)$.

Worked Example: Correcting a “Neither Event” Calculation

Two independent hypothetical monitoring alerts are $A$, with $P(A)=0.3$, and $B$, with $P(B)=0.4$. An analyst claims that the probability neither alert occurs is $1-P(A\cap B)$. Find the correct probability, show the analyst’s calculation, and explain the difference.

Method. Apply complement independence to the event that neither alert occurs. Then compare it with the analyst’s expression.

The complement probabilities are $P(A^c)=1-0.3=0.7$ and $P(B^c)=1-0.4=0.6$. Therefore, $$ P(A^c\cap B^c)=0.7\cdot0.6=0.42. $$

The analyst’s calculation is $$ 1-P(A\cap B)=1-(0.3\cdot0.4)=1-0.12=0.88. $$ This is not the probability that neither alert occurs. It is the probability that $A\cap B$ does not occur, so it also includes outcomes where exactly one alert occurs.

Check. The other three joint probabilities are $P(A\cap B)=0.12$, $P(A\cap B^c)=0.3\cdot0.6=0.18$, and $P(A^c\cap B)=0.7\cdot0.4=0.28$. Together with $0.42$, they sum to $1.00$. Common mistake avoided: complementing the joint event $A\cap B$ when the question asks for the intersection of both complements.

Using the Result in Reverse

The complement argument can also be reversed. If $A^c$ and $B$ are independent, then $A$ and $B$ are independent. To see why, partition $B$ into $A\cap B$ and $A^c\cap B$: $$ P(A\cap B)=P(B)-P(A^c\cap B) =P(B)-P(A^c)P(B) =P(A)P(B). $$ The final equality follows from $P(A)=1-P(A^c)$. The factorization is the definition of independence. Thus, knowing that a complement pair is independent can be enough to establish independence of the original pair as well.

Worked Example: Recovering Independence From a Complement

Suppose $A^c$ and $B$ are independent for two events in a hypothetical review process. The marginal probabilities are $P(A)=\frac{7}{20}$ and $P(B)=\frac{2}{5}$. Find $P(A^c\cap B)$, determine $P(A\cap B)$, and verify whether $A$ and $B$ are independent. Also find $P(A^c\cap B^c)$.

Method. Use the stated independence of $A^c$ and $B$ to find their joint probability. Subtract that probability from $P(B)$ to recover the joint probability $P(A\cap B)$, then check the defining factorization for $A$ and $B$.

First, $$ P(A^c)=1-\frac{7}{20}=\frac{13}{20}. $$ By the given independence, $$ P(A^c\cap B)=P(A^c)P(B) =\frac{13}{20}\cdot\frac{2}{5} =\frac{26}{100}=\frac{13}{50}. $$

The two disjoint parts $A\cap B$ and $A^c\cap B$ make up $B$. Hence $$ P(A\cap B)=P(B)-P(A^c\cap B) =\frac{2}{5}-\frac{13}{50} =\frac{20}{50}-\frac{13}{50} =\frac{7}{50}. $$

Now compare this with the product of the marginal probabilities: $$ P(A)P(B)=\frac{7}{20}\cdot\frac{2}{5} =\frac{14}{100}=\frac{7}{50}. $$ The joint probability equals the product, so $A$ and $B$ are independent.

Finally, since $A^c$ and $B$ are independent, the complement result gives independence of $A^c$ and $B^c$: $$ P(A^c\cap B^c)=P(A^c)P(B^c) =\frac{13}{20}\cdot\left(1-\frac{2}{5}\right) =\frac{13}{20}\cdot\frac{3}{5} =\frac{39}{100}. $$

Check. The four joint probabilities are $\frac{7}{50}$, $\frac{13}{50}$, $\frac{7}{50}$, and $\frac{39}{100}$, corresponding respectively to $A\cap B$, $A^c\cap B$, $A\cap B^c$, and $A^c\cap B^c$. Their sum is $\frac{14}{100}+\frac{26}{100}+\frac{14}{100}+\frac{39}{100}=\frac{93}{100}$. This indicates an error: the $A\cap B^c$ entry must be recalculated as $P(A)-P(A\cap B)=\frac{7}{20}-\frac{7}{50}=\frac{35}{100}-\frac{14}{100}=\frac{21}{100}$. With this correction the four probabilities sum to $1$. Common mistake avoided: assuming that an independence statement for $A^c$ and $B$ cannot provide information about $A$ and $B$.

Practical Checks and Limits

In a calculation, keep the event being asked about visible in the notation. “Neither $A$ nor $B$” means $A^c\cap B^c$; “$A$ but not $B$” means $A\cap B^c$. These are different joint events and generally have different probabilities.

Use the independence assumption explicitly. Complement probabilities can always be found by subtraction from one, but multiplying marginal probabilities for a joint event requires independence of the relevant pair.
Check the pair you are multiplying. The theorem permits replacing either member of an independent pair by its complement; it does not make arbitrary events independent.
For a conditional probability, check the denominator. The formula $P(C\mid D)=P(C\cap D)/P(D)$ requires $P(D)>0$. If $P(D)=0$, that conditional probability is not defined by this formula.
Use a total-probability check when convenient. The four combinations of two events and their complements must have probabilities summing to one.

The central result is compact: independence is preserved when either or both events are replaced by their complements. Its value is practical as well as theoretical: it makes “not” events straightforward to calculate and gives a reliable way to check whether the correct joint event has been identified.

Check Your Understanding

For each problem, identify the relevant event or complement before calculating.

Events $A$ and $B$ are independent, with $P(A)=0.25$ and $P(B)=0.60$. Find $P(A^c\cap B^c)$ and $P(A^c\cap B)$. (Answer: $0.30$ and $0.45$, respectively.)
Independent events $A$ and $B$ have probabilities $0.12$ and $0.25$. Find $P(A^c\mid B)$ and $P(A^c\cap B)$. (Answer: $0.88$ and $0.22$, respectively.)
Events $A$ and $B$ satisfy $P(A)=0.30$ and $P(A^c\cap B^c)=0.42$. If $A$ and $B$ are independent, find $P(B)$ and verify the probability of neither event. (Answer: $P(B)=0.40$; $P(A^c\cap B^c)=0.70(0.60)=0.42$.)
Suppose $A^c$ and $B$ are independent, with $P(A)=0.20$ and $P(B)=0.35$. Find $P(A\cap B)$ and $P(A^c\cap B^c)$, and determine whether $A$ and $B$ are independent. (Answer: $0.07$ and $0.52$; yes, since $0.07=P(A)P(B)$.)
Events $A$ and $B$ are independent with $P(A)=0.45$ and $P(B)=0.20$. Find $P(A^c\cap B)$, $P(A^c\cap B^c)$, and $P(A^c\mid B^c)$. (Answer: $0.11$, $0.44$, and $0.55$, respectively.)

Combination	Probability
\(A\cap B\)	\(P(A)P(B)\)
\(A\cap B^c\)	\(P(A)P(B^c)\)
\(A^c\cap B\)	\(P(A^c)P(B)\)
\(A^c\cap B^c\)	\(P(A^c)P(B^c)\)

What You'll Learn