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Statistics and Probability - Sets, including notation and operations up to three sets

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Set is a well-defined collection of distinct objects, called elements or members. We use notation x∈Ax \in A to indicate xx is an element of AA, and x∉Ax \notin A if it is not.

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The Universal Set (denoted by ξ\xi or UU) contains all possible elements under consideration in a particular context.

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The Empty Set (denoted by ∅\emptyset or {}\{\}) is a set containing no elements. Note that n(∅)=0n(\emptyset) = 0.

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Subsets: A⊆BA \subseteq B means every element of AA is also an element of BB. A⊂BA \subset B (Proper Subset) means AA is a subset of BB but A≠BA \neq B.

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Intersection (A∩BA \cap B): The set of elements that are in both AA AND BB. If A∩B=∅A \cap B = \emptyset, the sets are disjoint.

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Union (A∪BA \cup B): The set of elements that are in AA OR BB (or both).

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Complement (A′A' or AcA^c): The set of elements in the universal set ξ\xi that are NOT in set AA.

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Cardinality: The number of elements in a set AA, denoted as n(A)n(A).

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Venn Diagrams: Graphical representations used to show relationships between sets. For three sets A,B,A, B, and CC, the diagram consists of three overlapping circles within a rectangle representing ξ\xi.

📐Formulae

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) strips out the double-counted intersection.

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

n(A)+n(A′)=n(ξ)n(A) + n(A') = n(\xi) represents the total elements in the universal set.

n(A only)=n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C)n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)

💡Examples

Problem 1:

Given ξ={x:x∈Z,1≤x≤10}\xi = \{x : x \in \mathbb{Z}, 1 \le x \le 10\}, A={2,4,6,8,10}A = \{2, 4, 6, 8, 10\}, and B={1,2,3,4,5}B = \{1, 2, 3, 4, 5\}. Find: (i) A∩BA \cap B, (ii) A∪BA \cup B, (iii) (A∪B)′(A \cup B)'.

Solution:

(i) A∩B={2,4}A \cap B = \{2, 4\} (ii) A∪B={1,2,3,4,5,6,8,10}A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10\} (iii) Since ξ={1,2,3,4,5,6,7,8,9,10}\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, (A∪B)′=ξ∖(A∪B)={7,9}(A \cup B)' = \xi \setminus (A \cup B) = \{7, 9\}.

Explanation:

Intersection takes common elements. Union combines all elements from both sets without repetition. Complement finds elements in the universal set not present in the union.

Problem 2:

In a class of 3030 students, 1818 like Mathematics (MM), 1515 like Science (SS), and 88 like both. How many students like neither subject?

Solution:

Use the formula n(M∪S)=n(M)+n(S)−n(M∩S)n(M \cup S) = n(M) + n(S) - n(M \cap S). n(M∪S)=18+15−8=25n(M \cup S) = 18 + 15 - 8 = 25 Students who like neither = n(ξ)−n(M∪S)n(\xi) - n(M \cup S) 30−255\begin{array}{r} 30 \\ -25 \\ \hline 5 \end{array} So, 55 students like neither.

Explanation:

We first find the total number of students who like at least one subject using the inclusion-exclusion principle, then subtract that from the total class size.

Problem 3:

In a survey of 100100 people, 4040 read Magazine A, 3535 read B, and 3030 read C. 1515 read A and B, 1010 read B and C, 88 read A and C, and 33 read all three. Find the number of people who read at least one magazine.

Solution:

We use the three-set inclusion-exclusion formula: n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) n(A∪B∪C)=40+35+30−15−10−8+3n(A \cup B \cup C) = 40 + 35 + 30 - 15 - 10 - 8 + 3 n(A∪B∪C)=105−33+3=75n(A \cup B \cup C) = 105 - 33 + 3 = 75.

Explanation:

To find 'at least one', we calculate the union of all three sets. We add individual sets, subtract double-intersections, and add back the triple-intersection because it was subtracted one too many times.