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Statistics and Probability - Set Notation, Set-Builder Form, and Set Operations

Grade 8IB

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. If xx is an element of set AA, we write x∈Ax \in A. If it is not, we write x∉Ax \notin A.

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Roster Form: Listing all elements within curly brackets, e.g., A={2,4,6,8}A = \{2, 4, 6, 8\}.

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Set-Builder Form: Describing the properties that elements must satisfy using a variable, e.g., A={x∣x is an even natural number, x<10}A = \{x \mid x \text{ is an even natural number, } x < 10\}.

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Universal Set (ξ\xi or UU): The set containing all possible elements under consideration in a specific context.

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Empty Set (∅\emptyset or {}\{\} ): A set containing no elements.

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

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Intersection (A∩BA \cap B): The set of all elements that are common to both AA and BB.

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Complement (A′A'): The set of all elements in the universal set ξ\xi that are not in set AA.

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Cardinality (n(A)n(A)): The number of distinct elements in set AA.

📐Formulae

A∪B={x∣x∈A or x∈B}A \cup B = \{x \mid x \in A \text{ or } x \in B\}

A∩B={x∣x∈A and x∈B}A \cap B = \{x \mid x \in A \text{ and } x \in B\}

A′={x∣x∈ξ and x∉A}A' = \{x \mid x \in \xi \text{ and } x \notin A\}

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) (Inclusion-Exclusion Principle)

💡Examples

Problem 1:

Given the universal set ξ={x∣x∈Z,1≤x≤10}\xi = \{x \mid x \in \mathbb{Z}, 1 \leq x \leq 10\}. Let A={2,3,5,7}A = \{2, 3, 5, 7\} and B={x∣x is an even number in ξ}B = \{x \mid x \text{ is an even number in } \xi\}. Find A∩BA \cap B and n(A∪B)n(A \cup B).

Solution:

First, identify the elements: ξ={1,2,3,4,5,6,7,8,9,10}\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} A={2,3,5,7}A = \{2, 3, 5, 7\} B={2,4,6,8,10}B = \{2, 4, 6, 8, 10\}

Intersection: A∩B={2}A \cap B = \{2\}

For n(A∪B)n(A \cup B): A∪B={2,3,4,5,6,7,8,10}A \cup B = \{2, 3, 4, 5, 6, 7, 8, 10\} Counting the elements, n(A∪B)=8n(A \cup B) = 8.

Explanation:

We first converted the set-builder form of BB into roster form. A∩BA \cap B includes only the numbers present in both sets. A∪BA \cup B combines all elements, ensuring no duplicates are listed.

Problem 2:

If n(A)=15n(A) = 15, n(B)=12n(B) = 12, and n(A∩B)=7n(A \cap B) = 7, calculate n(A∪B)n(A \cup B).

Solution:

Using the formula: n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) n(A∪B)=15+12−7n(A \cup B) = 15 + 12 - 7 n(A∪B)=27−7n(A \cup B) = 27 - 7 n(A∪B)=20n(A \cup B) = 20

Explanation:

The Inclusion-Exclusion Principle is used here. We add the number of elements in AA and BB and subtract the overlapping elements (intersection) to avoid double-counting.

Problem 3:

Let ξ={a,b,c,d,e,f}\xi = \{a, b, c, d, e, f\} and S={a,e,i,o,u}∩ξS = \{a, e, i, o, u\} \cap \xi. Find S′S'.

Solution:

First, find SS: S={a,e,i,o,u}∩{a,b,c,d,e,f}S = \{a, e, i, o, u\} \cap \{a, b, c, d, e, f\} S={a,e}S = \{a, e\}

Now find the complement S′S' relative to ξ\xi: S′=ξ∖SS' = \xi \setminus S S′={b,c,d,f}S' = \{b, c, d, f\}

Explanation:

The intersection SS contains only elements found in both the vowel set and the universal set. The complement S′S' consists of all elements in ξ\xi that are not in SS.