4.2

Probability Models

  • A description of a random phenomenon in the language of mathematics is called a probability model.

Components of a Probability Model

  • A probability model consists of:
    • A list of all possible outcomes.
    • A probability for each outcome.

Sample Space

  • A sample space is the set of all outcomes of a random phenomenon.
  • The symbol typically employed to represent a sample space is the letter S.

Examples of Sample Spaces

Simple Sample Spaces

  • The sample space for the toss of a coin is:
    S=heads,tailsS = {\text{heads}, \text{tails}}
  • The sample space for the roll of a die is:
    S=1,2,3,4,5,6S = {1, 2, 3, 4, 5, 6}

More Complex Sample Spaces

  • The sample space for the number of heads in 4 tosses of a coin is:
    S=0,1,2,3,4S = {0, 1, 2, 3, 4}
  • The sample space for the systolic pressure of an adult female in units of mm Hg is roughly:
    S=x  ∣  70≤x≤340S = {x \;|\; 70 \leq x \leq 340}

Finding a Sample Space

Example

  • Problem: Find the sample space of outcomes for 4 coin tosses.

Solution

  • The sample space consists of the following outcomes:
    • HHHH
    • HHHT
    • HHTH
    • HTHH
    • THHH
    • HHTT
    • HTHT
    • THHT
    • TTHH
    • THTH
    • HTTH
    • HTTT
    • THTT
    • TTHT
    • TTTH
    • TTTT

Listing Specific Events from Sample Space

Example (Finding a Specific Event)

  • Problem: List all outcomes for the event “exactly 2 heads.”
  • Denote this event by the letter E.

Solution

  • The event E is represented as follows:
    E=HHTT, HTHT, HTTH, THHT, THTH, TTHHE = {\text{HHTT, HTHT, HTTH, THHT, THTH, TTHH}}

Probability Rules

  • The probability P(A)P(A) of any event AA satisfies the conditions:
    0≤P(A)≤10 \leq P(A) \leq 1
  • If SS is the sample space, then:
    P(S)=1P(S) = 1
  • Two events AA and BB are disjoint if they have no outcomes in common. If AA and BB are disjoint, then:
    P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
  • The complement of an event AA is the event AcA^c that AA does not occur, with
    P(Ac)=1−P(A)P(A^c) = 1 - P(A)

Characteristics of Random Phenomena

  • For every random phenomenon, there exists an event with probability zero, known as the null event, denoted by the mathematical symbol for the empty set:
    ∅\emptyset
  • Because every random phenomenon must result in some outcome, the null event is also called the impossible event. It’s worth noting that the complement of the sample space SS is:
    ∅\emptyset

Example of Finding Probabilities

Example

  • Problem: Consider the sample space of outcomes for 4 coin tosses. Suppose every outcome is equally likely. Find the probabilities of events:
    • A=2 or fewer headsA = {2 \text{ or fewer heads}}
    • B=exactly 3 headsB = {\text{exactly 3 heads}}

Solution

  • The outcomes previously listed:
    • HHHH
    • HHHT
    • HHTH
    • HTHH
    • THHH
    • HHTT
    • HTHT
    • THHT
    • TTHH
    • THTH
    • HTTH
    • HTTT
    • THTT
    • TTHT
    • TTTH
    • TTTT

Outcomes Count

  • To calculate the probabilities:
    P(A)=count of outcomes in Acount of outcomes in SP(A) = \frac{\text{count of outcomes in A}}{\text{count of outcomes in S}}
    P(B)=count of outcomes in Bcount of outcomes in SP(B) = \frac{\text{count of outcomes in B}}{\text{count of outcomes in S}}
  • Outcomes in event A: 11
  • Total outcomes in the sample space S: 16
    P(A)=1116P(A) = \frac{11}{16}
  • Outcomes in event B: 4
    P(B)=416P(B) = \frac{4}{16}

Visualizing Probabilities

  • A Venn Diagram can be used to visualize the probabilities of events A and B.

Example of Complements and Event Calculations

Example (Finding Probability of a Complement)

  • Problem: Find the probabilities of event C=3 or more headsC = {3 \text{ or more heads}} by using the definition of event A and the complement rule.

Solution

  • To find P(C)P(C): P(C)=1−P(A)P(C) = 1 - P(A)
    • Here, this results in:
      P(C)=P(Ac)=1−P(A)P(C) = P(A^c) = 1 - P(A)
  • Numerically this is structured as:
    P(C)=1−1116=516P(C) = 1 - \frac{11}{16} = \frac{5}{16}

Visual Representation of Event Probabilities

  • This answer can also be visualized using a Venn Diagram.

Multiplication Rule for Independent Events

  • Two events AA and BB are independent if knowing that one occurs does not change the probability that the other occurs.
  • If events AA and BB are independent, then:
    P(A and B)=P(A)⋅P(B)P(A \text{ and } B) = P(A) \cdot P(B)

Example of Independent Events

  • Scenario: Consider tossing a fair coin successively two times.
    • Let events
      A=head on first tossA = {\text{head on first toss}}
      B=head on second tossB = {\text{head on second toss}}
    • Find the probability of P(A and B)P(A \text{ and } B).

Solution

  • Because the events are independent, we have:
    P(A and B)=P(A)⋅P(B)=12⋅12=14P(A \text{ and } B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

Example of Multiple Coin Tosses

Example

  • Problem: Consider tossing a fair coin successively four times. Let events
    Ai=head on i-th tossA_i = {\text{head on i-th toss}}
  • Find the probability of getting 4 heads.

Solution

  • The probability calculated is:
    P(A<em>1)=P(A</em>2)=P(A<em>3)=P(A</em>4)=12P(A<em>1) = P(A</em>2) = P(A<em>3) = P(A</em>4) = \frac{1}{2}
  • Therefore:
    P(A)=P(A<em>1)⋅P(A</em>2)⋅P(A<em>3)⋅P(A</em>4)=(12)4=116P(A) = P(A<em>1) \cdot P(A</em>2) \cdot P(A<em>3) \cdot P(A</em>4) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}

Example from Genetics

Genetic Experiment Case Study

  • Example: Gregor Mendel used garden peas in some of his genetic experiments.
    • These experiments revealed that inheritance operates randomly.
    • Seeds can either be green (G) or yellow (Y); the yellow gene is dominant.

Problem Statement

  • Find the probability that the offspring are green.

Solution

  • The probability is evaluated as follows: P(G)=P(G and G)=P(G)⋅P(G)P(G) = P(G \text{ and } G) = P(G) \cdot P(G)
    • Assuming independent genes:
      P(G)=12⋅12=14P(G) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}