Week 11 · Operations on events, mutually exclusive events and the probability of an event when the outcomes are equally likely · 3 lessons
1
Union, intersection, complement and difference of events, and De Morgan’s laws
Statistics Sec 3 Term 1, Unit 3, Lesson 1
GoalCombine events with ∪, ∩, ′ and −, say each one in words, and use De Morgan’s laws.
New words
union A ∪ B · A or B or both occur: at least one of themintersection A ∩ B · both A and B occurcomplement A′ · A does not occurdifference A − B · A occurs but B does not; it equals A ∩ B′
Explanation
An event is a set of outcomes, so events can be combined like sets, and every combination has a plain meaning. Take a spinner with twelve equal sectors numbered 1 to 12, so S = {1, 2, …, 12}, and let A = “even” = {2, 4, 6, 8, 10, 12} and B = “a multiple of 3” = {3, 6, 9, 12}. The intersection A ∩ B = {6, 12} holds the outcomes that are in both sets: A and B occur together. The union A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12} holds the outcomes that are in at least one set. The complement A′ = {1, 3, 5, 7, 9, 11} is everything in S that is not in A, so A ∪ A′ = S and A ∩ A′ = ∅. The difference A − B = {2, 4, 8, 10} means “A but not B”. De Morgan’s laws turn a “not” inside out: the outcomes in neither A nor B form (A ∪ B)′ = A′ ∩ B′, and the outcomes that are not in both form (A ∩ B)′ = A′ ∪ B′. In words, “A and B do not both occur” is the same as “at most one of them occurs”. Do not mix it up with “neither”, which is the stronger statement. The picture shows the four regions of this spinner.
Worked examples
Using the spinner above, list B′, B − A and (A ∪ B)′, and check that (A ∪ B)′ = A′ ∩ B′. We get B′ = {1, 2, 4, 5, 7, 8, 10, 11}, B − A = {3, 9} and (A ∪ B)′ = {1, 5, 7, 11}. On the other side A′ = {1, 3, 5, 7, 9, 11}, and A′ and B′ have exactly the outcomes 1, 5, 7, 11 in common, so A′ ∩ B′ is the same set.
Phrase
Symbol
rain and a late bus
R ∩ L
rain or a late bus (or both)
R ∪ L
rain but the bus is on time
R − L
neither rain nor a late bus
(R ∪ L)′ = R′ ∩ L′
not both together (at most one)
(R ∩ L)′ = R′ ∪ L′
From words to symbols. On a school day let R = “it rains” and L = “the bus is late”. The table gives each phrase its symbol.
The second law on the same spinner. Let C = “greater than 8” = {9, 10, 11, 12} and D = “odd” = {1, 3, 5, 7, 9, 11}. Then C ∩ D = {9, 11}, so (C ∩ D)′ has ten outcomes: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12. On the other side C′ = {1, …, 8} and D′ = {2, 4, 6, 8, 10, 12}, and their union is the same ten outcomes. So (C ∩ D)′ = C′ ∪ D′.
Class activity
Write the numbers 1 to 10 on ten slips. Let A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8, 10}. Draw two overlapping circles on a sheet and place each slip in its region. Read off A ∩ B, A ∪ B, A − B and (A ∪ B)′, then write down A′ and B′ and check that A′ ∩ B′ gives the same slips as (A ∪ B)′.
Practice
1
S = {a, b, c, d, e, f, g, h}, A = {a, b, c, d} and B = {c, d, e, f}. Write A ∩ B, A ∪ B and A − B.
A ∩ B = {c, d}; A ∪ B = {a, b, c, d, e, f}; A − B = {a, b}.
2
With the same sets find B′, (A ∪ B)′ and A′ ∩ B′. What do you notice?
B′ = {a, b, g, h}; (A ∪ B)′ = {g, h}; A′ = {e, f, g, h}, so A′ ∩ B′ = {g, h}. The last two sets are equal: De Morgan’s first law.
3
In a school let L = “the pupil has a library card” and M = “the pupil has a canteen card”. Write with symbols: (a) a library card but no canteen card; (b) neither card; (c) the pupil does not have both cards.
(a) L − M = L ∩ M′; (b) (L ∪ M)′ = L′ ∩ M′; (c) (L ∩ M)′ = L′ ∪ M′.
4
Why do A and A′ always satisfy A ∩ A′ = ∅ and A ∪ A′ = S?
A′ holds exactly the outcomes of S that are not in A. So no outcome is in both sets (A ∩ A′ = ∅), and every outcome is in one of the two (A ∪ A′ = S).
5
On the twelve-sector spinner let E = {4, 8, 12} and F = “even”. Find E ∩ F, E ∪ F and E − F. What is special about E and F?
E ∩ F = {4, 8, 12} = E; E ∪ F = {2, 4, 6, 8, 10, 12} = F; E − F = ∅. Every outcome of E is in F: E is a subset of F.
