Grade by Grade
A. Counting and Probability · Probability

Events as Sets and the Complement of an Event

A1.5About 65 minutes
Learning goals
  • I can describe an event as a set of outcomes and as a subset of the sample space, using set notation.
  • I can find the complement of an event and use P(A′) = 1 − P(A).
  • I can use the complement to solve "at least one" problems efficiently.

Words to know

event
A set of outcomes; a subset of the sample space. We write A ⊆ S.
simple event
An event with exactly one outcome, such as rolling a 4.
subset (⊆)
A ⊆ S means every outcome in A is also in S.
empty set (∅)
The set with no outcomes: an impossible event, P(∅) = 0.
complement (A′)
The event that A does not happen: every outcome of S that is not in A. Also written Ā or ~A ("not A").
element (∈)
x ∈ A means the outcome x is in the set A.

Events are sets

An event is any collection of outcomes, so it is a subset of the sample space. Using set notation for a die roll:

  • S = {1, 2, 3, 4, 5, 6}
  • A = "roll a prime number" = {2, 3, 5}, so A ⊆ S and n(A) = 3
  • B = "roll a 4" = {4} is a simple event (one outcome)
  • C = "roll a 9" = ∅, the empty set, an impossible event
  • S itself is the event "roll something", a certain event

We write 3 ∈ A ("3 is an element of A") and 4 ∉ A. Thinking of events as sets lets you use set operations (complement, intersection, union) to build new events. Those operations are the language of the next three lessons.

Remember from Grade 8: you used Venn diagrams and the words and, or and not with probabilities. The symbols in this unit are the standard ones you will see in university courses.

The complement

The complement of A, written A′ (also Ā or ~A), is the event "A does not happen". It contains every outcome in S that is not in A.

For A = {2, 3, 5}: A′ = {1, 4, 6}.

Every outcome is in A or in A′, never both. So

n(A) + n(A′) = n(S) and P(A) + P(A′) = 1, which gives P(A′) = 1 − P(A).

This simple rule is one of the most useful in the course. When an event is complicated, its complement is often simple.

SA{2, 3, 5}primeA′1 4 6Die roll: S = {1, 2, 3, 4, 5, 6}A = prime = {2, 3, 5}A′ = not prime = {1, 4, 6}n(A) + n(A′) = n(S)P(A) + P(A′) = 1
The complement A′ is everything in the sample space S outside A. Together, A and A′ cover S with no overlap, so their probabilities add to 1.
Worked example

Example 1: Set notation

One card is drawn from a standard deck. Let F = "face card" and H = "heart". (a) Is the king of hearts ∈ F? (b) Describe F′ in words and find n(F′). (c) Find P(F′) two ways.

  1. (a) The king of hearts is a face card, so K♥ ∈ F. (It is also ∈ H.)
  2. (b) F′ = "not a face card": the aces and the 2 to 10 of every suit. n(F′) = 52 − 12 = 40.
  3. (c) Counting: P(F′) = 40/52 = 10/13.
  4. Complement rule: P(F′) = 1 − P(F) = 1 − 12/52 = 1 − 3/13 = 10/13. ✓

(a) Yes. (b) Not a face card, n(F′) = 40. (c) P(F′) = 10/13.

The "at least one" strategy

"At least one" problems have many cases (exactly 1, exactly 2, exactly 3, …). The complement of "at least one" is "none", which is just one case:

P(at least one) = 1 − P(none).

When trials are independent, P(none) is found by multiplying (you multiplied probabilities of independent events in Grade 8, and lesson 10 develops this fully).

second diefirst die11622633644655666666666At least one 6: 11 cellsNo 6 (the complement): 5 × 5 = 25P(at least one 6)= 1 − P(no 6)= 1 − 25/36 = 11/36
Rolling two dice: 11 of the 36 outcomes contain at least one 6. It is quicker to count the 25 outcomes with no 6 (5 choices for each die) and subtract.
Worked example

Example 2: At least one 6 with two dice

Two dice are rolled. Find P(at least one 6).

  1. Complement: "no 6 on either die". Each die has 5 non-six faces, so there are 5 × 5 = 25 such outcomes.
  2. P(no 6) = 25/36.
  3. P(at least one 6) = 1 − 25/36 = 11/36.
  4. Check by direct counting: row 6 has 6 cells, column 6 has 6 cells, and (6, 6) is in both, so 6 + 6 − 1 = 11. ✓

P(at least one 6) = 11/36 ≈ 0.306.

Worked example

Example 3: At least one 6 in four rolls

A die is rolled 4 times. Find P(at least one 6).

  1. Complement: no 6 in any of the 4 rolls.
  2. Each roll has P(not 6) = 5/6, and the rolls are independent, so P(no 6) = (5/6)⁴ = 625/1296.
  3. P(at least one 6) = 1 − 625/1296 = 671/1296.

P(at least one 6) = 671/1296 ≈ 0.518, just better than even.

Worked example

Example 4: At least one defective

Each smoke detector in a batch has a 3% chance of being defective, independently. A store sells a pack of 10. Find P(at least one detector in the pack is defective).

  1. P(a detector works) = 1 − 0.03 = 0.97.
  2. P(all 10 work) = 0.97¹⁰ ≈ 0.7374.
  3. P(at least one defective) = 1 − 0.97¹⁰ ≈ 1 − 0.7374 = 0.2626.

About 0.263, or roughly 1 pack in 4. That is why each detector should be tested after installation.

Spot the complement

Watch for the phrases at least one, not, none, some, any. "At least one" ↔ "none". "Not all" ↔ "all". "At most 4" ↔ "5 or more". If the direct way has many cases, try the complement.

Just for fun

Modern probability theory grew out of gambling questions. In the 1650s the French writer Antoine Gombaud, known as the Chevalier de Méré, asked Blaise Pascal about dice bets like the one in Example 3. Pascal's letters with Pierre de Fermat about such problems are often called the birth of probability theory. One unhappy gambler, one whole branch of mathematics.

Hands-on activity

Count the complement

You need: a deck of cards and two dice.

  1. Without counting card by card, use the complement to find how many cards are not clubs, not aces, and not red face cards. Then check one of them by sorting the deck.
  2. Roll two dice 36 times and record whether at least one 6 appears. Compare your experimental probability with 11/36 ≈ 0.31.
  3. Challenge a family member: "Roll one die 4 times. I bet at least one 6 appears." Play 20 rounds (no money) and keep score. Who wins more often?

What to look for: not clubs 39, not aces 48, not red face cards 46. In step 2 expect roughly 11 of 36 rolls with a 6, though anywhere from about 6 to 16 is normal. In step 3 the bettor on "at least one 6" wins about 52% of the time, so 20 rounds may go either way.

Your turn

1

S = {1, 2, …, 10}. A = {multiples of 3}. List A and A′.

2

P(A) = 0.37. Find P(A′).

3

One card is drawn. Find P(not a spade).

4

Five coins are tossed. Find P(at least one head).

5

Two dice are rolled. Find P(the sum is not 7).

6

A die is rolled 3 times. Find P(at least one 6).

7

Write the complement of "at most 2 heads" in 5 coin tosses.

More practice: warm-up

1

P(snow) = 0.8. P(no snow)?

2

S = {a, b, c, d, e}, A = {a, e}. A′?

3

What is the complement of S?

4

n(S) = 52 and n(A) = 4. n(A′)?

5

What is P(∅)?

More practice: on your own

1

Three dice are rolled. Find P(at least one 1).

2

A test has 6 true-or-false questions answered by guessing. Find P(at least one correct).

3

A family has 4 children; assume each is equally likely to be a girl or a boy, independently. Find P(at least one girl).

4

A ferry is late on 8% of its trips, independently. Find P(it is late at least once in 5 trips).

5

Explain why P(A) + P(A′) = 1 for every event A.

Challenge

1

The Chevalier de Méré also bet on getting at least one double six in 24 rolls of two dice. Find this probability and compare it with Example 3.

2

How many times must a die be rolled so that P(at least one 6) exceeds 0.9?

3

Each of 3 independent alarm systems detects a break-in with probability 0.85. Find P(at least one detects it). Why do buildings use backup systems?

Check your understanding

Quiz

1.

S = {1, 2, 3, 4, 5, 6, 7, 8}. A = {even numbers}. What is A′?

2.

P(B) = 5/12. What is P(B′)?

3.

A coin is tossed 6 times. What is P(at least one tail)?

4.

Which is the complement of "at least 2 of the 5 phones are defective"?

5.

A die is rolled twice. What is P(at least one 6)?

6.

Which event is the empty set when one die is rolled?

For parents and teachers

Big idea (A1.5, events and complements): an event is a set of outcomes. Its complement A′ is everything else, and P(A′) = 1 − P(A). For "at least one" problems, the fastest route is usually 1 − P(none).

How to help: Ask "what's the opposite?" questions: the opposite of "at least one head" is "no heads"; the opposite of "at most 2" is "3 or more". Try the de Méré bet in the activity together.

Common difficulties:

  • Adding probabilities for "at least one" (1/6 + 1/6 for two dice), which double-counts.
  • Getting the complement of "at least" and "at most" wrong by one.
  • Writing P(A′) = 1/P(A) instead of 1 − P(A).
  • Unfamiliar symbols: ⊆ (subset), ∈ (element of), ∅ (empty set), A′ (complement).