Independence of Complements and Unions

Why Complements and Unions Need Care

Knowing that two events are independent lets us calculate probabilities involving either event’s complement. It does not, by itself, let us treat a union of several events as independent of another event. The distinction matters whenever a calculation combines events: independence is a precise probability condition, not a rule that automatically passes to every combination.

We will first establish the complement rules, then identify conditions under which a union calculation is justified. Throughout, events are subsets of a probability space, and \(A^c\) denotes the complement of \(A\). Events \(A\) and \(B\) are independent when \(P(A\cap B)=P(A)P(B)\).

Complements of Independent Events

Theorem: If \(A\) and \(B\) are independent, then each of the pairs \((A^c,B)\), \((A,B^c)\), and \((A^c,B^c)\) is independent. In particular, \(A\) is independent of \(B\) if and only if \(A^c\) is independent of \(B\). The same equivalence holds when complementing \(B\), or both events.

The proof uses only the definition of independence and the identities for probabilities of complements. For example,

P(A^c\cap B) =P(B)-P(A\cap B) =P(B)-P(A)P(B) =P(A^c)P(B).

Thus \(A^c\) and \(B\) are independent. The other cases follow by the same calculation. Explicitly,

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

and

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

Each equality uses either a complement identity or the assumed independence of \(A\) and \(B\). The reverse implications also follow: for instance, if \(A^c\) and \(B\) are independent, then \(P(A\cap B)=P(B)-P(A^c\cap B)=P(B)(1-P(A^c))=P(A)P(B)\). \(\square\)

Worked Example: Probabilities After Taking Complements

Suppose \(A\) and \(B\) are independent, with \(P(A)=0.30\) and \(P(B)=0.40\). Find the probabilities of \(A\cap B\), \(A^c\cap B\), \(A\cap B^c\), and \(A^c\cap B^c\).

Independence gives \(P(A\cap B)=0.30(0.40)=0.12\). The marginal probabilities of the complements are \(P(A^c)=0.70\) and \(P(B^c)=0.60\). By the complement theorem,

\begin{aligned} P(A^c\cap B)&=0.70(0.40)=0.28,\\ P(A\cap B^c)&=0.30(0.60)=0.18,\\ P(A^c\cap B^c)&=0.70(0.60)=0.42. \end{aligned}

The four probabilities sum to \(0.12+0.28+0.18+0.42=1\), as they must: the four intersections partition the sample space. This check also catches arithmetic errors.

Unions of Disjoint Events

The union \(B\cup C\) occurs when at least one of \(B\) or \(C\) occurs. If \(B\) and \(C\) are disjoint, they cannot occur together, so their probabilities add. If they overlap, the overlap must be subtracted once to avoid counting it twice.

Union formula: For any two events \(B\) and \(C\), \(P(B\cup C)=P(B)+P(C)-P(B\cap C)\). If \(B\cap C=\varnothing\), this reduces to \(P(B\cup C)=P(B)+P(C)\).

Suppose \(A\) is independent of each of two disjoint events \(B\) and \(C\). Then \(A\) is independent of their union:

\begin{aligned} P\bigl(A\cap(B\cup C)\bigr) &=P\bigl((A\cap B)\cup(A\cap C)\bigr)\\ &=P(A\cap B)+P(A\cap C)\\ &=P(A)P(B)+P(A)P(C)\\ &=P(A)P(B\cup C). \end{aligned}

The second equality requires disjointness: \(A\cap B\) and \(A\cap C\) are disjoint when \(B\) and \(C\) are. Without that condition, the calculation needs an overlap term.

Worked Example: Combining Disjoint Events

An equally likely outcome is selected from the 12 ordered pairs \((i,j)\), where \(i\in\{0,1,2\}\) and \(j\in\{1,2,3,4\}\). Let \(A\) be the event \(i\in\{0,1\}\), let \(B\) be \(j=1\), and let \(C\) be \(j=2\). Find \(P(B\cup C)\) and check whether \(A\) is independent of this union.

There are 8 outcomes in \(A\), so \(P(A)=8/12=2/3\). Each of \(B\) and \(C\) contains 3 outcomes, and \(A\cap B\) and \(A\cap C\) each contain 2. Thus \(P(B)=P(C)=3/12=1/4\) and \(P(A\cap B)=P(A\cap C)=2/12=1/6\). These equalities verify that \(A\) is independent of both \(B\) and \(C\).

The events \(B\) and \(C\) are disjoint, so \(P(B\cup C)=1/4+1/4=1/2\). Also, \(A\cap(B\cup C)\) contains 4 outcomes, giving probability \(4/12=1/3\). The product is \(P(A)P(B\cup C)=(2/3)(1/2)=1/3\). Hence \(A\) is independent of \(B\cup C\), as the disjoint-union result predicts.

When a Union Is Not Independent

For overlapping events, inclusion-exclusion gives the exact criterion:

\begin{aligned} P\bigl(A\cap(B\cup C)\bigr) &=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C),\\ P(A)P(B\cup C) &=P(A)\bigl(P(B)+P(C)-P(B\cap C)\bigr). \end{aligned}

Therefore, \(A\) is independent of \(B\cup C\) precisely when the two right-hand sides are equal. Independence of \(A\) with \(B\) and with \(C\) supplies only the first two terms of the first expression. It does not generally determine \(P(A\cap B\cap C)\), nor does it guarantee that this intersection equals \(P(A)P(B\cap C)\).

A useful sufficient condition is that \(A\) be independent of \(B\), \(C\), and \(B\cap C\). Substituting those three factorizations into the first expression gives \(P(A)[P(B)+P(C)-P(B\cap C)]=P(A)P(B\cup C)\). Disjointness is another sufficient condition because then \(B\cap C=\varnothing\) and the overlap terms vanish. In particular, mutual independence of \(A,B,C\) is sufficient: it gives the needed factorizations, including \(P(A\cap B\cap C)=P(A)P(B)P(C)\) and \(P(B\cap C)=P(B)P(C)\).

Worked Example: Pairwise Independence Does Not Protect a Union

Choose \(X\) and \(Y\) independently, each taking values 0 and 1 with probability \(1/2\). Let \(A=\{X=1\}\), \(B=\{Y=1\}\), and \(C=\{X\ne Y\}\). Check whether \(A\) is independent of \(B\) and \(C\), and then check whether it is independent of \(B\cup C\).

The four equally likely outcomes are \((0,0),(0,1),(1,0),(1,1)\). Each of \(A,B,C\) has probability \(1/2\). The event \(A\cap B\) contains only \((1,1)\), and \(A\cap C\) contains only \((1,0)\). Each intersection has probability \(1/4=(1/2)(1/2)\), so \(A\) is independent of \(B\) and of \(C\).

The union \(B\cup C\) contains \((0,1),(1,0),(1,1)\), so its probability is \(3/4\). Every outcome in \(A\) belongs to this union, hence \(P(A\cap(B\cup C))=P(A)=1/2\). But \(P(A)P(B\cup C)=(1/2)(3/4)=3/8\), which is not \(1/2\). Thus \(A\) is not independent of \(B\cup C\).

This example also shows why checking pairs one at a time is not enough: \(A\) is independent of each component, but the combined event carries information about \(A\).

Unions of Several Independent Events

For two independent events, inclusion-exclusion immediately yields \(P(B\cup C)=P(B)+P(C)-P(B)P(C)\). Equivalently, use complements: \((B\cup C)^c=B^c\cap C^c\), and the complements are independent. For more than two events, the product formula requires mutual independence, not merely pairwise independence.

Theorem: If \(A_1,\ldots,A_m\) are mutually independent events, meaning that for every nonempty index set \(I\subseteq\{1,\ldots,m\}\), \(P(\bigcap_{i\in I}A_i)=\prod_{i\in I}P(A_i)\), then \(P(\bigcup_{i=1}^m A_i)=1-\prod_{i=1}^m(1-P(A_i))\).

To prove the result, take complements and use De Morgan’s law: \((\bigcup_{i=1}^m A_i)^c=\bigcap_{i=1}^m A_i^c\). Mutual independence is preserved when events are replaced by complements. Indeed, for disjoint index sets \(I,J\), the probability of any chosen intersection of original events and complements is

\begin{aligned} P\left(\bigcap_{i\in I}A_i\cap\bigcap_{j\in J}A_j^c\right) &=\sum_{K\subseteq J}(-1)^{|K|} P\left(\bigcap_{i\in I\cup K}A_i\right)\\ &=\sum_{K\subseteq J}(-1)^{|K|} \prod_{i\in I\cup K}P(A_i)\\ &=\prod_{i\in I}P(A_i)\prod_{j\in J}(1-P(A_j)). \end{aligned}

The first equality is inclusion-exclusion applied to the complements indexed by \(J\); the second uses mutual independence. Taking \(I=\varnothing\) and \(J=\{1,\ldots,m\}\) gives \(P(\bigcap_i A_i^c)=\prod_i(1-P(A_i))\). Taking the complement proves the stated union formula. \(\square\)

Worked Example: Three Mutually Independent Events

Suppose \(A_1,A_2,A_3\) are mutually independent, with probabilities \(0.20\), \(0.50\), and \(0.10\). Find the probability that at least one occurs.

The probability that none occurs is the probability of the intersection of their complements:

P(A_1^c\cap A_2^c\cap A_3^c) =(1-0.20)(1-0.50)(1-0.10) =(0.80)(0.50)(0.90)=0.36.

Therefore \(P(A_1\cup A_2\cup A_3)=1-0.36=0.64\). The calculation depends on mutual independence, which justifies multiplying the three complement probabilities.

Common Mistake: Multiplying Without the Needed Independence

A tempting but incorrect argument is: “\(A\) is independent of \(B\), and \(A\) is independent of \(C\), so \(A\) is independent of \(B\cup C\).” The example with \(X,Y\) shows the error. Individual factorizations do not supply the intersection probability needed by inclusion-exclusion.

Likewise, pairwise independence among several events does not justify the product formula for the probability that none occurs. For the same \(A,B,C\) example, each pair is independent, but \(P(A\cup B\cup C)=3/4\): the only outcome outside the union is \((0,0)\). The formula \(1-(1-1/2)^3=7/8\) is wrong here because the events are not mutually independent.

Before applying an independence calculation, identify exactly which intersections the calculation uses. Complementing an event preserves its independence relationships. Combining events into a union requires additional structure, such as disjointness or the relevant mutual-independence conditions.

Check Your Understanding

Answer each question by identifying the probability identity or independence condition that justifies your calculation.

If \(A\) and \(B\) are independent with probabilities \(0.6\) and \(0.25\), find \(P(A^c\cap B^c)\).
If \(B\) and \(C\) are disjoint and each is independent of \(A\), what condition ensures that \(A\) is independent of \(B\cup C\)? Explain the calculation.
Why does independence of \(A\) from each of \(B\) and \(C\) not generally imply independence of \(A\) from \(B\cup C\)?
For three events, what independence condition justifies \(P(A_1\cup A_2\cup A_3)=1-\prod_{i=1}^3(1-P(A_i))\)?
State the inclusion-exclusion expression for \(P(A\cap(B\cup C))\), and identify the overlap term that must be accounted for.