Random Events and the Sample Space

Random Experiments and the Sample Space

Definition of a Random Experiment

Random Experiment

An experiment that can be repeated under the same conditions and whose outcome cannot be determined in advance before each trial is called a random experiment.

Characteristics of a Random Experiment

Characteristics:

  1. Repeatability: the experiment can be repeated under the same conditions
  2. Uncertainty: the outcome of each trial cannot be determined in advance
  3. Observability: the outcome of the experiment can be observed

Definition of the Sample Space

Sample Space

The set consisting of all possible outcomes of a random experiment is called the sample space, denoted by Ω\Omega.

Definition of a Sample Point

Sample Point

Each element of the sample space is called a sample point, denoted by ω\omega.

Examples

Example 1: the random experiment of tossing a coin

  • Sample space: Ω={heads,tails}\Omega = \{\text{heads}, \text{tails}\}
  • Sample points: ω1=heads\omega_1 = \text{heads}, ω2=tails\omega_2 = \text{tails}

Example 2: the random experiment of rolling a die

  • Sample space: Ω={1,2,3,4,5,6}\Omega = \{1, 2, 3, 4, 5, 6\}
  • Sample points: ωi=i\omega_i = i, i=1,2,…,6i = 1, 2, \dots, 6

Example 3: the random experiment of measuring the length of an object

  • Sample space: Ω=[0,+∞)\Omega = [0, +\infty)
  • Sample points: ω∈[0,+∞)\omega \in [0, +\infty)

Random Events

Definition of a Random Event

Random Event

A subset of the sample space is called a random event, or simply an event.

Classification of Events

Classification:

  1. Certain event: the sample space Ω\Omega itself
  2. Impossible event: the empty set ∅\varnothing
  3. Elementary event: an event containing exactly one sample point
  4. Compound event: an event containing several sample points

Representations of an Event

Representations:

  1. Set representation: A={ω1,ω2,…,ωn}A = \{\omega_1, \omega_2, \dots, \omega_n\}
  2. Descriptive representation: A={heads appears}A = \{\text{heads appears}\}
  3. Conditional representation: A={ω:ω satisfies some condition}A = \{\omega : \omega \text{ satisfies some condition}\}

Examples

Example 4: rolling a die and defining events

  • A={an even number appears}={2,4,6}A = \{\text{an even number appears}\} = \{2, 4, 6\}
  • B={a number greater than 3 appears}={4,5,6}B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}
  • C={a 1 appears}={1}C = \{\text{a 1 appears}\} = \{1\} (an elementary event)

Relations and Operations of Events

Containment of Events

Containment of Events

If the occurrence of event AA necessarily implies the occurrence of event BB, then AA is said to be contained in BB, denoted by A⊂BA \subset B.

Properties:

  • A⊂AA \subset A (reflexivity)
  • If A⊂BA \subset B and B⊂AB \subset A, then A=BA = B (antisymmetry)
  • If A⊂BA \subset B and B⊂CB \subset C, then A⊂CA \subset C (transitivity)

Equality of Events

Equality of Events

If A⊂BA \subset B and B⊂AB \subset A, then events AA and BB are said to be equal, denoted by A=BA = B.

Union of Events

Union of Events

The union of events AA and BB, denoted by A∪BA \cup B, is the event that AA occurs or BB occurs.

Properties:

  • A∪A=AA \cup A = A (idempotence)
  • A∪B=B∪AA \cup B = B \cup A (commutativity)
  • (A∪B)∪C=A∪(B∪C)(A \cup B) \cup C = A \cup (B \cup C) (associativity)

Intersection of Events

Intersection of Events

The intersection of events AA and BB, denoted by A∩BA \cap B, is the event that AA occurs and BB occurs.

Properties:

  • A∩A=AA \cap A = A (idempotence)
  • A∩B=B∩AA \cap B = B \cap A (commutativity)
  • (A∩B)∩C=A∩(B∩C)(A \cap B) \cap C = A \cap (B \cap C) (associativity)

Difference of Events

Difference of Events

The difference of events AA and BB, denoted by A−BA - B or A∖BA \setminus B, is the event that AA occurs but BB does not.

Complement Event

Complement Event

The complement event of event AA, denoted by A‾\overline{A} or AcA^c, is the event that AA does not occur.

Properties:

  • A∩A‾=∅A \cap \overline{A} = \varnothing
  • A∪A‾=ΩA \cup \overline{A} = \Omega
  • A‾‾=A\overline{\overline{A}} = A

Mutually Exclusive Events

Mutually Exclusive Events

If A∩B=∅A \cap B = \varnothing, then events AA and BB are said to be mutually exclusive (or incompatible).

Operation Laws of Events

Distributive laws:

  • A∩(B∪C)=(A∩B)∪(A∩C)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
  • A∪(B∩C)=(A∪B)∩(A∪C)A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

De Morgan’s laws:

  • A∪B‾=A‾∩B‾\overline{A \cup B} = \overline{A} \cap \overline{B}
  • A∩B‾=A‾∪B‾\overline{A \cap B} = \overline{A} \cup \overline{B}

Examples

Example 5: the die-rolling experiment

Let Ω={1,2,3,4,5,6}\Omega = \{1, 2, 3, 4, 5, 6\}, and define the events:

  • A={an even number appears}={2,4,6}A = \{\text{an even number appears}\} = \{2, 4, 6\}
  • B={a number greater than 3 appears}={4,5,6}B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}
  • C={a 1 appears}={1}C = \{\text{a 1 appears}\} = \{1\}

Computation:

  • A∪B={2,4,5,6}A \cup B = \{2, 4, 5, 6\}
  • A∩B={4,6}A \cap B = \{4, 6\}
  • A−B={2}A - B = \{2\}
  • A‾={1,3,5}\overline{A} = \{1, 3, 5\}
  • AA and CC are mutually exclusive

Example 6: the coin-tossing experiment

Let Ω={heads,tails}\Omega = \{\text{heads}, \text{tails}\}, and define the events:

  • A={heads appears}={heads}A = \{\text{heads appears}\} = \{\text{heads}\}
  • B={tails appears}={tails}B = \{\text{tails appears}\} = \{\text{tails}\}

Properties:

  • AA and BB are mutually exclusive
  • A∪B=ΩA \cup B = \Omega
  • A=B‾A = \overline{B}, B=A‾B = \overline{A}

Exercises

Exercise 1

Roll a die and define the events A={an even number appears}A = \{\text{an even number appears}\}, B={a number greater than 3 appears}B = \{\text{a number greater than 3 appears}\}. Find A∪BA \cup B and A∩BA \cap B.

event operationsample spaceset operation

Approach: Compute from the definitions of events and the rules of event operations.

Steps:

  1. A={an even number appears}={2,4,6}A = \{\text{an even number appears}\} = \{2, 4, 6\}
  2. B={a number greater than 3 appears}={4,5,6}B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}
  3. A∪B={2,4,5,6}A \cup B = \{2, 4, 5, 6\}
  4. A∩B={4,6}A \cap B = \{4, 6\}

Answer: A∪B={2,4,5,6}A \cup B = \{2, 4, 5, 6\}, A∩B={4,6}A \cap B = \{4, 6\}

Exercise 2

Let Ω={1,2,3,4,5}\Omega = \{1, 2, 3, 4, 5\}, A={1,2,3}A = \{1, 2, 3\}, B={2,3,4}B = \{2, 3, 4\}. Find A‾\overline{A} and A−BA - B.

event operationcomplement eventdifference event

Approach: Compute from the definitions of the complement event and the difference event.

Steps:

  1. A‾=Ω−A={4,5}\overline{A} = \Omega - A = \{4, 5\}
  2. A−B=A∩B‾={1,2,3}∩{1,5}={1}A - B = A \cap \overline{B} = \{1, 2, 3\} \cap \{1, 5\} = \{1\}

Answer: A‾={4,5}\overline{A} = \{4, 5\}, A−B={1}A - B = \{1\}

Exercise 3

Prove De Morgan’s law: A∪B‾=A‾∩B‾\overline{A \cup B} = \overline{A} \cap \overline{B}.

event operationDe Morgan's lawproof

Approach: Prove De Morgan’s law through set operations.

Steps:

  1. Let ω∈A∪B‾\omega \in \overline{A \cup B}; then ω∉A∪B\omega \notin A \cup B
  2. Therefore ω∉A\omega \notin A and ω∉B\omega \notin B
  3. That is, ω∈A‾\omega \in \overline{A} and ω∈B‾\omega \in \overline{B}
  4. Therefore ω∈A‾∩B‾\omega \in \overline{A} \cap \overline{B}
  5. Conversely, let ω∈A‾∩B‾\omega \in \overline{A} \cap \overline{B}; then ω∈A‾\omega \in \overline{A} and ω∈B‾\omega \in \overline{B}
  6. Therefore ω∉A\omega \notin A and ω∉B\omega \notin B
  7. That is, ω∉A∪B\omega \notin A \cup B, so ω∈A∪B‾\omega \in \overline{A \cup B}

Answer: the proof is complete

Exercise 4

Let AA and BB be arbitrary events. Prove that: (A∪B)∩A‾=B∩A‾(A \cup B) \cap \overline{A} = B \cap \overline{A}.

event operationdistributive lawproof

Approach: Use the distributive law and the properties of event operations.

Steps:

  1. (A∪B)∩A‾=(A∩A‾)∪(B∩A‾)(A \cup B) \cap \overline{A} = (A \cap \overline{A}) \cup (B \cap \overline{A})
  2. A∩A‾=∅A \cap \overline{A} = \varnothing
  3. Therefore (A∪B)∩A‾=∅∪(B∩A‾)=B∩A‾(A \cup B) \cap \overline{A} = \varnothing \cup (B \cap \overline{A}) = B \cap \overline{A}

Answer: the proof is complete

Exercise 5

Let Ω={a,b,c,d}\Omega = \{a, b, c, d\}, A={a,b}A = \{a, b\}, B={b,c}B = \{b, c\}, C={c,d}C = \{c, d\}. Find (A∪B)∩C(A \cup B) \cap C.

event operationorder of operationsset operation

Approach: Compute following the order of event operations.

Steps:

  1. A∪B={a,b,c}A \cup B = \{a, b, c\}
  2. (A∪B)∩C={a,b,c}∩{c,d}={c}(A \cup B) \cap C = \{a, b, c\} \cap \{c, d\} = \{c\}

Answer: (A∪B)∩C={c}(A \cup B) \cap C = \{c\}

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