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Math 7 · S.CP.A.1 · 7.SP.C.7

Complements and probabilities that add to one

A forecast gives the chance of rain tomorrow. What is the chance that it stays dry, and why is that number already settled by the first one?

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Why this exists

A forecast gives a 36\% chance of rain tomorrow. Nobody prints the chance of no rain beside it, and nobody needs to. Tomorrow will either have rain or not, never both, so the two chances share the whole of the certainty between them. Once one number is known, the other is fixed.

The same idea saves work. “At least one head” in a toss of several coins covers many outcomes, while its opposite, “no heads”, is a single one. Count whichever side is smaller.

This lesson names the event “A does not happen”, gives it notation, and shows why its probability and the probability of A always add to 1. Underneath sits a second fact: the probabilities of all the outcomes in a sample space add to 1.

Definition

Start with an experiment, its sample space S, and an event A, which is a collection of outcomes from S.

The complement of A is the event made of every outcome in S that is not in A. It happens exactly when A does not happen.

This lesson writes the complement as \text{not } A. Other books write A' (read “A prime”), A^c (read “A complement”) or \overline{A} (read “A bar”). All four name the same event.

For one spin of the prize wheel in Figure 1, the eight equal sections are the outcomes. Let M be the event “lands on miss”, the three shaded sections. Then \text{not } M is “lands on pen, cap or mug”, the five unshaded sections.

miss pen cap miss mug pen miss cap
Figure 1Eight equal sections. You win a prize unless the arrow stops in one of the three shaded “miss” sections.

The second idea of the lesson is about the probabilities themselves. Each outcome of a sample space has a probability from 0 to 1, and the probabilities of all the outcomes add to 1. A list of numbers that does not total 1 cannot be the probabilities of a complete sample space.

Intuition

Think of the sample space as a whole that is cut into two pieces. Everything on one side of the cut is A; everything on the other side is \text{not } A. Nothing is left over and nothing sits on both sides. Figure 2 shows this for two coins tossed together: the four outcomes split into “no heads”, which is only \text{TT}, and “at least one head”, which is the other three.

HH HT TH TT at least one head no heads
Figure 2The sample space for two coins, cut into an event and its complement. Every outcome is on exactly one side.

Probability measures how much of the whole each piece holds, on a scale where the whole sample space, the certain event, counts as 1. Two pieces that make up the whole hold all of it between them, so their probabilities add to 1. When one piece is small the other is large: a 36\% chance of rain leaves 64\% for no rain.

Results

Complement. For any event A in a sample space S, the events A and \text{not } A share no outcome, since an outcome cannot be both in A and outside it, and together they contain every outcome of S.

Complement rule. For any event A,

P(A) + P(\text{not } A) = 1

and so

P(\text{not } A) = 1 - P(A)

With percents the whole is 100\%, so P(\text{not } A) = 100\% - P(A).

All outcomes together. If the outcomes of S are o_1, o_2, \dots, o_n, then

P(o_1) + P(o_2) + \cdots + P(o_n) = 1

Complement of a complement. The complement of \text{not } A is A itself.

Why it is true

All outcomes together. Each time the experiment runs, exactly one outcome happens: at least one, because S lists every possible result, and no more than one, because the outcomes of a sample space do not overlap. So the event S is certain, and P(S) = 1. The probability of an event is the sum of the probabilities of its outcomes, so all the outcomes of S add to 1.

The complement rule, equally likely outcomes. Suppose S has n equally likely outcomes and A contains k of them. Every outcome not in A is in \text{not } A, so \text{not } A contains the other n - k. Using probability as favorable outcomes over all outcomes,

\begin{aligned} P(A) + P(\text{not } A) &= \frac{k}{n} + \frac{n - k}{n}\\ &= \frac{n}{n} = 1 \end{aligned}

The complement rule, any model. When the outcomes are not equally likely, add the probabilities of the outcomes in A to get P(A), and those of the outcomes in \text{not } A to get P(\text{not } A). Every outcome of S is in exactly one of the two events, so adding P(A) and P(\text{not } A) adds each outcome’s probability exactly once. That is the total over all of S, which is 1.

Subtracting P(A) from both sides of P(A) + P(\text{not } A) = 1 gives P(\text{not } A) = 1 - P(A); subtracting P(\text{not } A) instead gives P(A) = 1 - P(\text{not } A).

Complement of a complement. The outcomes not in \text{not } A are the ones that are not outside A, which are exactly the outcomes in A.

Worked examples

Example 1

Rain and no rain

(a) A forecast gives a 36\% chance of rain in town tomorrow. What is the chance of no rain?

(b) Kerenza’s weather app shows only the chance of no rain for Sunday, 0.86. What is the chance of rain on Sunday?

Solution

(a) “No rain tomorrow” is the complement of “rain tomorrow”: tomorrow either has rain or does not, and cannot do both. Working in percents, where the whole is 100\%,

P(\text{no rain}) = 100\% - 36\% = 64\%

As a decimal this is 1 - 0.36 = 0.64.

(b) This time the number given belongs to the complement, so the rule is used the other way round:

\begin{aligned} P(\text{rain}) &= 1 - P(\text{no rain})\\ &= 1 - 0.86 = 0.14 \end{aligned}

Check. 0.14 + 0.86 = 1, as the two must.

Example 2

Win unless it lands on miss

On the prize wheel of Figure 1, Henrietta spins once and wins a prize unless the arrow stops on miss. Find the probability that she wins.

Solution

The eight sections are equal, so the eight outcomes are equally likely. Let M be “lands on miss”. Three of the eight sections are miss, so

P(M) = \frac{3}{8}

Henrietta wins exactly when M does not happen, so the event “she wins” is \text{not } M:

\begin{aligned} P(\text{win}) &= P(\text{not } M)\\ &= 1 - \frac{3}{8} = \frac{5}{8} \end{aligned}

Check. Counting the winning sections directly: two pen, two cap and one mug, which is 2 + 2 + 1 = 5 of the 8, or \tfrac{5}{8} again.

Example 3

A vending machine with three outcomes

When the oat bar button on a vending machine is pressed, one of three things happens: the bar drops, the “sold out” light comes on, or the machine jams and gives nothing. Suppose the bar drops with probability 0.87 and the sold-out light comes on with probability 0.08.

(a) Find the probability that the machine jams.

(b) Find the probability that the bar is not sold out, in two different ways.

Solution

These three outcomes are not equally likely, so there is nothing to count; the probabilities are given instead.

(a) The three outcomes make up the whole sample space, so their probabilities add to 1:

0.87 + 0.08 + P(\text{jams}) = 1

That is 0.95 + P(\text{jams}) = 1, and subtracting 0.95 from both sides gives P(\text{jams}) = 0.05.

(b) First way, by the complement rule:

P(\text{not sold out}) = 1 - 0.08 = 0.92

Second way, by listing the outcomes in the event. “Not sold out” is the set \{\text{drops}, \text{jams}\}, so its probability is 0.87 + 0.05 = 0.92. The two ways agree.

Example 4

A seed that sprouts four times as often as it fails

Ignatia plants a seed. Suppose the probability that it sprouts is four times the probability that it does not sprout. Find both probabilities.

Solution

Let p be the probability that the seed does not sprout. Then the probability that it sprouts is 4p. “Sprouts” and “does not sprout” are complements, so their probabilities add to 1:

4p + p = 1

Collecting like terms gives 5p = 1, and dividing both sides by 5,

p = \frac{1}{5} = 0.2

So the seed fails to sprout with probability 0.2 and sprouts with probability 4 \times 0.2 = 0.8.

Check. 0.8 is four times 0.2, and 0.8 + 0.2 = 1.

Example 5

Which events are complements?

Use one spin of the prize wheel in Figure 1.

(a) Let B be “lands on pen or cap”. Describe \text{not } B and find its probability.

(b) Are “lands on miss” and “lands on pen” complements of each other?

(c) What is the complement of \text{not } M, where M is “lands on miss”?

Solution

(a) B holds the two pen and two cap sections, 4 in all. The sections left over are the three miss sections and the mug, so \text{not } B is “lands on miss or mug”. By the complement rule,

P(\text{not } B) = 1 - \frac{4}{8} = \frac{4}{8} = \frac{1}{2}

Counting directly gives the same: 3 + 1 = 4 sections of 8.

(b) No. Being complements takes two things: sharing no outcome, and together holding every outcome of S. These two events pass the first test, since the arrow stops in one section only, but fail the second: the cap and mug sections are in neither. The probabilities show the gap, since \tfrac{3}{8} + \tfrac{2}{8} = \tfrac{5}{8}, not 1, and the missing \tfrac{3}{8} belongs to cap and mug. The complement of “lands on miss” is “lands on pen, cap or mug”.

(c) \text{not } M is the pen, cap and mug sections. The outcomes not in it are the miss sections, so the complement of \text{not } M is M.

Example 6

At least one head

(a) Gwilym-Rhys tosses two fair coins. Use the complement to find the probability of at least one head.

(b) Lysandra tosses three fair coins. Find the probability of at least one head.

Solution

(a) Listing the sample space with the first coin’s result first gives \{\text{HH}, \text{HT}, \text{TH}, \text{TT}\}, four equally likely outcomes. “At least one head” means one head or two heads. Its complement is “no heads”, and the only outcome with no heads is \text{TT}, as Figure 2 shows. So

\begin{aligned} P(\text{at least one head}) &= 1 - P(\text{no heads})\\ &= 1 - \frac{1}{4} = \frac{3}{4} \end{aligned}

Check. Counting directly, \text{HH}, \text{HT} and \text{TH} each have a head: 3 of 4.

(b) Fixing the first coin, then the second, then the third, the sample space is

\begin{aligned} &\text{HHH},\ \text{HHT},\ \text{HTH},\ \text{HTT},\\ &\text{THH},\ \text{THT},\ \text{TTH},\ \text{TTT} \end{aligned}

which is 8 equally likely outcomes. “At least one head” holds seven of them, but its complement, “no heads”, is the single outcome \text{TTT}:

P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}

Each extra coin makes the “at least one head” side bigger, while “no heads” stays one outcome.

Boundaries

The sample space has to be fixed first. “Not A” means “the rest of S”, so it changes when S changes. On a wheel with an extra “book” section, “not miss” would include book. With no sample space stated, a complement is not yet defined.

Opposite-sounding words are not always complements. A complement must share no outcome with the event and must also cover every other outcome. On a number cube, “more than 3” and “less than 3” share nothing, but both leave out 3. “Rain” and “sunny” fail both ways: a dry, cloudy day is in neither, and a sunny shower is in both.

Two events that share nothing need not add to 1. Their probabilities add to 1 only when together they fill S; otherwise the sum falls short, as with miss and pen in Example 5.

Two events that fill S need not add to 1 either. If they overlap, the shared outcomes are counted twice. For two coins, “at least one head” and “at least one tail” between them cover all four outcomes, but \tfrac{3}{4} + \tfrac{3}{4} = \tfrac{3}{2}, because \text{HT} and \text{TH} are in both.

The outcomes must be listed completely, each once. A list that misses a result, or lists one twice, can total something other than 1 without any number in it being wrong.

Where this goes next

A probability model attaches a probability to every outcome, and those probabilities adding to 1 is the first thing any model must satisfy; the same check runs on every probability distribution table. Odds set an event’s share against its complement’s share. And with compound events, where listing everything stops being practical, “none at all” usually stays simple, so 1 - P(\text{none}) remains the way in.

Exercises

Naming the complement

  1. For one spin of the prize wheel in Figure 1, describe the complement of “lands on cap” in words, and find its probability.
  2. Two coins are tossed, with sample space \{\text{HH}, \text{HT}, \text{TH}, \text{TT}\}. List the complement of “both coins show the same face”.
  3. Kerenza says the complement of “the team wins the match” is “the team loses the match”. Is she right? Explain.
  4. In the two-coin toss, let A = \{\text{HH}\}. List A^c, then list the complement of A^c.

Using the complement rule

  1. A forecast gives a 27\% chance of rain. Give the chance of no rain as a percent and as a decimal.
  2. The probability that a seed of a certain kind sprouts is 0.94. What is the probability that it does not sprout?
  3. For another kind of seed, the probability that a seed does not sprout is \tfrac{2}{15}. What is the probability that it sprouts?
  4. For the prize wheel in Figure 1, find P(\text{not pen}).
  5. The wheel’s rules change: Jonah now wins unless the arrow stops on miss or on mug. What is the probability that he wins?
  6. The probability that a chosen snack in a vending machine is sold out is 0.035. What is the probability that it is not sold out?

All the outcomes together

  1. On a vending machine’s crisps button, suppose the crisps drop with probability 0.91 and the machine jams with probability 0.02. The only other outcome is the sold-out light. Find the probability of the sold-out light.
  2. A forecast splits tomorrow into three outcomes: no rain with probability 0.4, light rain with probability 0.35, and heavy rain. (a) Find P(\text{heavy rain}). (b) Find the probability of rain of any kind, in two different ways.
  3. Henrietta proposes these probabilities for the season a chosen person’s birthday falls in: spring 0.3, summer 0.25, autumn 0.25, winter 0.3. Can these be the probabilities of the four outcomes? Explain.
  4. Each morning Oleander takes the bus, takes the tram, cycles or walks. The bus and the tram are equally likely, P(\text{cycles}) = 0.2 and P(\text{walks}) = 0.4. Find P(\text{bus}).

Counting the other side

  1. Two fair coins are tossed. Use the complement to find the probability of at least one tail.
  2. Three fair coins are tossed, with the sample space listed in Example 6. Find (a) the probability of at least one tail, and (b) the probability that the three coins do not all show the same face.
  3. Ignatia tosses two fair coins and wins if she gets at least one head and at least one tail. Use the complement to find the probability that she wins.

Reasoning

  1. Lysandra writes P(A) = 0.6 and P(\text{not } A) = 0.5 for the same event A. Explain why this cannot be right.
  2. An event is exactly as likely as its complement. Show that its probability must be \tfrac{1}{2}.
  3. Use the complement rule to explain why no event can have a probability greater than 1.
Answers
  1. “Lands on miss, pen or mug”. Two of the eight sections are cap, so the probability is 1 - \tfrac{2}{8} = \tfrac{6}{8} = \tfrac{3}{4}.
  2. \{\text{HT}, \text{TH}\}. The outcomes with the same face are \text{HH} and \text{TT}; the complement is the rest.
  3. No, if a match can end in a draw. The complement of “wins” is “does not win”, which holds both a loss and a draw. “Loses” shares no outcome with “wins”, but it leaves out the draw, so the two together do not fill the sample space.
  4. A^c = \{\text{HT}, \text{TH}, \text{TT}\}, and the complement of A^c is \{\text{HH}\}, which is A again.
  5. 73\%, or 0.73. 100\% - 27\% = 73\%.
  6. 0.06. 1 - 0.94 = 0.06.
  7. \tfrac{13}{15}. 1 - \tfrac{2}{15} = \tfrac{13}{15}.
  8. \tfrac{3}{4}. Two of the eight sections are pen, so 1 - \tfrac{2}{8} = \tfrac{6}{8} = \tfrac{3}{4}.
  9. \tfrac{1}{2}. He loses on 3 miss sections and 1 mug, 4 of 8, so he wins with probability 1 - \tfrac{4}{8} = \tfrac{1}{2}.
  10. 0.965. 1 - 0.035 = 0.965.
  11. 0.07. The three probabilities add to 1, so the sold-out light has probability 1 - 0.91 - 0.02 = 0.07.
  12. (a) 0.25, since 1 - 0.4 - 0.35 = 0.25. (b) 0.6. Adding the two kinds of rain, 0.35 + 0.25 = 0.6; or, since rain of any kind is the complement of no rain, 1 - 0.4 = 0.6.
  13. No. 0.3 + 0.25 + 0.25 + 0.3 = 1.1, which is more than 1. The probabilities of all the outcomes must add to exactly 1.
  14. 0.2. Writing x for P(\text{bus}), which is also P(\text{tram}), the four add to 1: 2x + 0.2 + 0.4 = 1, so 2x = 0.4 and x = 0.2.
  15. \tfrac{3}{4}. The complement, no tails, is \{\text{HH}\}, so the probability is 1 - \tfrac{1}{4} = \tfrac{3}{4}.
  16. (a) \tfrac{7}{8}. The complement, no tails, is only \text{HHH}, so 1 - \tfrac{1}{8} = \tfrac{7}{8}. (b) \tfrac{3}{4}. The complement, all the same, is \{\text{HHH}, \text{TTT}\}, so 1 - \tfrac{2}{8} = \tfrac{6}{8} = \tfrac{3}{4}.
  17. \tfrac{1}{2}. She loses when both coins match, \{\text{HH}, \text{TT}\}, which is 2 of 4, so she wins with probability 1 - \tfrac{2}{4} = \tfrac{1}{2}. Directly, \text{HT} and \text{TH} give the same \tfrac{2}{4}.
  18. The two add to 0.6 + 0.5 = 1.1. Each time, exactly one of A and \text{not } A happens, so their probabilities must add to exactly 1; at least one of the two numbers is wrong.
  19. Let p = P(A). Then P(\text{not } A) = p as well, and the complement rule gives p + p = 1. So 2p = 1 and p = \tfrac{1}{2}.
  20. By the complement rule P(A) = 1 - P(\text{not } A), and P(\text{not } A) is at least 0 because no probability is negative. Taking a number that is at least 0 away from 1 leaves at most 1, so P(A) \le 1, with equality only when P(\text{not } A) = 0.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0