The Complement Rule, introduction to probability lesson

A batch of three indicator lamps is checked, and the question is the probability that at least one lamp is faulty. That event gathers exactly one, exactly two and three faults, while its opposite, no faulty lamp, is a single case. This lesson proves that in every probability model an event has probability one minus the probability of its complement, and shows how to find the correct complement for phrases such as at least one, not all and at least k.

Key terms

Complementary partition

Definition The complement of an event $A$ is the event $A^{c}$ of all outcomes in $\Omega$ that are not in $A$ . The two events are disjoint and together exhaustive,

\begin{gathered} A \cap A^{c} = \varnothing, \\[0.4em] A \cup A^{c} = \Omega, \end{gathered}

so when $A$ is neither $\varnothing$ nor $\Omega$ they form a partition of $\Omega$ into two nonempty events. Conversely, an event $B$ is the complement of $A$ exactly when $A \cap B = \varnothing$ and $A \cup B = \Omega$ , that is, when $B$ is disjoint from $A$ and the two events together are exhaustive.

In plain words Every outcome lies in exactly one of $A$ and $A^{c}$ . A proposed opposite event is the complement exactly when it shares no outcome with $A$ and leaves out no outcome outside $A$ , so both conditions have to be checked.

Example A delivery forecast uses the event $A$ that a package arrives by its deadline. The event that it arrives exactly one minute late is disjoint from $A$ , but it misses every other late arrival, so it is not $A^{c}$ . If the model also allows the package never to arrive, then even “arrives after the deadline” leaves outcomes out. The complement is “does not arrive by the deadline”, which contains every late arrival and nondelivery. If the forecast gives $A$ probability $0.7$ , the remaining $0.3$ belongs to all of these outcomes together, not to the one-minute case alone.

Complement rule

Definition For every event $A$ of a probability model,

P(A^{c}) = 1 - P(A),

and equivalently $P(A) = 1 - P(A^{c})$ . The proof uses only the axioms. The events $A$ and $A^{c}$ are disjoint with union $\Omega$ , so additivity and normalization give $P(A) + P(A^{c}) = P(\Omega) = 1$ . Nothing is assumed about independence, equally likely outcomes, the number of outcomes or repeated trials.

In plain words The probability inside an event and the probability outside it share the total of one. When the opposite event is easier to describe or to compute, find its probability and subtract from one. Most mistakes happen in naming the opposite event, not in the subtraction.

Example A constructed model for a batch of three lamps records a pattern such as GFG, with G for good and F for faulty in the positions of lamps 1, 2 and 3. The model gives every pattern a probability.

Faults	Patterns	Probability of each	Group total
$0$	GGG	$0.700$	$0.700$
$1$	FGG, GFG, GGF	$0.080$	$0.240$
$2$	FFG, FGF, GFF	$0.015$	$0.045$
$3$	FFF	$0.015$	$0.015$

The probabilities are nonnegative and the group totals add to $1$ , so the table is a valid model. The event that at least one lamp is faulty is the complement of $\{\text{GGG}\}$ , so its probability is $1 - 0.7 = 0.3$ . Adding the seven patterns with a fault gives the same value, $0.24 + 0.045 + 0.015 = 0.3$ , with more work.

At least one

Definition For events $A_1, \dots, A_n$ , the event that at least one of them occurs is the union $A_1 \cup \dots \cup A_n$ . By De Morgan’s law its complement is the event that none of them occurs,

\begin{aligned} &(A_1 \cup \dots \cup A_n)^{c} \\[0.4em] &\quad = A_1^{c} \cap \dots \cap A_n^{c}, \end{aligned}

so the complement rule gives

\begin{aligned} &P(\text{at least one}) \\[0.4em] &\quad = 1 - P(\text{none}). \end{aligned}

In plain words “At least one” gathers many cases, exactly one, exactly two and so on, while “none” is often a single case. The probability of none still has to come from the model. It is the probability of an intersection, and replacing it by a product of separate probabilities needs a reason such as independence, which the model has to supply.

Example In the lamp table let $F_i$ be the event that lamp $i$ is faulty. The event $F_1 \cup F_2 \cup F_3$ has complement $F_1^{c} \cap F_2^{c} \cap F_3^{c} = \{\text{GGG}\}$ , which gives the probability $0.3$ found above. Lamp 1 is faulty in FGG and in the three patterns FFG, FGF and FFF, which have probability $0.015$ each, so

\begin{aligned} P(F_1) &= 0.080 + 3(0.015) \\[0.4em] &= 0.125, \end{aligned}

and the same holds for lamps 2 and 3. Multiplying the three probabilities of a good lamp gives $(0.875)^3 = 343/512 \approx 0.670$ , not the stated $0.7$ for GGG. So this table does not describe lamps that fail independently, and the product would give the wrong value $1 - 343/512 \approx 0.330$ for at least one fault. The complement rule needed no independence, because it used the stated probability of the whole complementary event.

Not all

Definition For events $A_1, \dots, A_n$ , the event that not all of them occur is the complement of $A_1 \cap \dots \cap A_n$ . By De Morgan’s law

\begin{aligned} &(A_1 \cap \dots \cap A_n)^{c} \\[0.4em] &\quad = A_1^{c} \cup \dots \cup A_n^{c}, \end{aligned}

which is the event that at least one of them fails to occur. The complement rule gives

P(\text{not all}) = 1 - P(\text{all}).

In plain words “Not all” is a weak denial. It holds as soon as one event fails, while “none” needs every event to fail.

Example In the lamp table, “not all lamps are good” is the complement of $\{\text{GGG}\}$ , so its probability is $1 - 0.7 = 0.3$ . It is the same event as “at least one lamp is faulty”. The event “no lamp is good” is $\{\text{FFF}\}$ , with probability $0.015$ . Reading “not all good” as “all not good” would replace $0.3$ by $0.015$ .

Complement of at least k

Definition Let $N$ be a count attached to each outcome that takes only integer values, such as the number of faulty lamps, and let $k$ be an integer. Write $\{N \ge k\}$ for the event of all outcomes whose count is at least $k$ . Then

\begin{aligned} \{N \ge k\}^{c} &= \{N < k\} \\[0.4em] &= \{N \le k - 1\}, \\[0.4em] \{N > k\}^{c} &= \{N \le k\}, \end{aligned}

so $P(N \ge k) = 1 - P(N \le k - 1)$ and $P(N > k) = 1 - P(N \le k)$ . For a quantity that can take values between integers, the complement of “at least $k$ ” is still “less than $k$ ”, but it need not be the same as “at most $k - 1$ ”.

In plain words The complement of “at least $k$ ” is “fewer than $k$ ”. The boundary value $k$ belongs to “at least $k$ ”, so it has to leave the complement, and turning $\ge$ into $\le$ without that adjustment counts the boundary on both sides.

Example A constructed model for the number $N$ of defective items in a sample of four assigns these probabilities.

$n$	$0$	$1$	$2$	$3$	$4$
$P(N = n)$	$0.50$	$0.25$	$0.15$	$0.07$	$0.03$

The complement of $\{N \ge 2\}$ is $\{N \le 1\}$ , with probability $0.75$ , so $P(N \ge 2) = 0.25$ . Direct addition agrees, since $0.15 + 0.07 + 0.03 = 0.25$ . The event $\{N > 2\}$ has complement $\{N \le 2\}$ , with probability $0.90$ , so $P(N > 2) = 0.10$ . Taking $\{N \le 2\}$ as the complement of $\{N \ge 2\}$ would give $0.10$ for at least two defects, losing the $0.15$ on exactly two. Taking “no defects” as the complement would give $0.50$ , because exactly one defect would then be counted as at least two.

Check yourself

Using the lamp table, find the probability of at most one fault. Then use it to find the probability of at least two faults, and name the complement you use.

Show the answer

With $N$ the number of faulty lamps, at most one fault is $\{N \le 1\}$ , the pattern GGG together with the three one-fault patterns, so its probability is $0.7 + 0.24 = 0.94$ . For a whole-number count, a value that is not at most $1$ is at least $2$ , so the complement of $\{N \le 1\}$ is $\{N \ge 2\}$ . Hence the probability of at least two faults is $1 - 0.94 = 0.06$ . Direct addition confirms it, since $0.045 + 0.015 = 0.06$ .

Using the lamp table, find the probability that not all lamps are faulty and the probability that at least one lamp is good. Then explain why “not all faulty” and “all good” are different events.

Show the answer

Not all faulty is the complement of $\{\text{FFF}\}$ , so its probability is $1 - 0.015 = 0.985$ . A pattern has at least one good lamp exactly when it is not FFF, so at least one good lamp is the same event with the same probability $0.985$ .

All good contains only GGG and has probability $0.7$ . The pattern GFF is not all faulty and has a good lamp, but it is not all good. That one outcome shows that the two events differ.

A second batch has a different joint table, and the only information given is that all three lamps are good with probability $0.84$ . Which of these probabilities can you find, at least one fault, not all good, exactly one fault, all faulty?

Show the answer

At least one fault is the complement of all good, so its probability is $1 - 0.84 = 0.16$ . Not all good is the same event, so it also has probability $0.16$ .

Exactly one fault and all faulty are not determined. The remaining $0.16$ could sit entirely on the one-fault patterns, or it could be split as $0.10$ on exactly one fault, $0.04$ on exactly two and $0.02$ on all three. Both splits give valid models with all good at $0.84$ , yet exactly one fault gets $0.16$ in the first and $0.10$ in the second, and all faulty gets $0$ and $0.02$ . Filling the gap by assuming independent lamps would add an assumption that the given information does not contain.

The Complement Rule

Key terms

Complementary partition

Complement rule

At least one

Not all

Complement of at least k

Check yourself

How this lesson connects

Builds on

Used later in