Chapter 5 – Probability: An Introduction to Modeling Uncertainty

5.1 Events and Probabilities

  • Random Experiment: A process that generates well-defined outcomes.
  • Sample Space: The set of all possible outcomes of a random experiment.
    • Example 1: Tossing a coin.
      • Outcomes: Head, Tail.
      • Sample space: S=Head,TailS = {Head, Tail}
    • Example 2: Rolling a die.
      • Outcomes: 1, 2, 3, 4, 5, 6.
      • Sample space: S=1,2,3,4,5,6S = {1, 2, 3, 4, 5, 6}
  • Event: A collection of outcomes.
    • Example: CP&L Capacity Expansion Project
      • Possible project durations: 8, 9, 10, 11, or 12 months.
      • Event CC: Project completed in 10 months or less =8,9,10= {8, 9, 10}
  • Probability of an Event: The sum of probabilities of outcomes for the event.
    • P(C)=P(8)+P(9)+P(10)=0.15+0.25+0.30=0.70P(C) = P(8) + P(9) + P(10) = 0.15 + 0.25 + 0.30 = 0.70
    • P(L)P(L): Project completed in less than 10 months =8,9= {8, 9}
      • P(L)=P(8)+P(9)=0.15+0.25=0.40P(L) = P(8) + P(9) = 0.15 + 0.25 = 0.40
    • P(M)P(M): Project completed in more than 10 months =11,12= {11, 12}
      • P(M)=P(11)+P(12)=0.15+0.15=0.30P(M) = P(11) + P(12) = 0.15 + 0.15 = 0.30

5.2 Some Basic Relationships of Probability

  • Complement of an Event: The event consisting of all outcomes not in AA, denoted by AcA^c.
    • P(A)+P(Ac)=1P(A) + P(A^c) = 1
    • P(A)=1P(Ac)P(A) = 1 - P(A^c)
  • Union of Two Events: The event containing all outcomes in AA or BB or both, denoted by ABA \cup B.
  • Intersection of Two Events: The event containing all outcomes belonging to both AA and BB, denoted by ABA \cap B.
  • Addition Law: Computes the probability of the union of two events.
    • P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B)
    • Rationale: Outcomes in ABA \cap B are counted twice when computing P(A)+P(B)P(A) + P(B), so we subtract P(AB)P(A \cap B) to correct for overcounting.
  • Application of the Addition Law: HR study example.
    • Event SS: Dissatisfaction with salary, P(S)=0.30P(S) = 0.30
    • Event WW: Dissatisfaction with work assignments, P(W)=0.20P(W) = 0.20
    • Event SWS \cap W: Dissatisfaction with both salary and work assignments, P(SW)=0.12P(S \cap W) = 0.12
    • Probability of leaving due to dissatisfaction with salary or work assignments:
      • P(SW)=P(S)+P(W)P(SW)=0.30+0.200.12=0.38P(S \cup W) = P(S) + P(W) - P(S \cap W) = 0.30 + 0.20 - 0.12 = 0.38
  • Mutually Exclusive Events: Two events with no outcomes in common. If one occurs, the other cannot.
    • P(AB)=0P(A \cap B) = 0
    • For mutually exclusive events, the addition law simplifies to: P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

5.3 Conditional Probability

  • Studying mortgage defaults at Lancaster Savings and Loans (S&L).

  • SS = event that a customer is single.

  • MM = event that a customer is married.

  • DD = event that a customer defaulted on their mortgage.

  • DcD^c = event that a customer did not default on their mortgage.

  • Joint Probabilities: Intersection probabilities of events.

    • Probability of not defaulting and being married: P(DcM)=64300=0.2133P(D^c \cap M) = \frac{64}{300} = 0.2133
    • Probability of defaulting and being married: P(DM)=79300=0.2633P(D \cap M) = \frac{79}{300} = 0.2633
    • Probability of not defaulting and being single: P(DcS)=116300=0.3867P(D^c \cap S) = \frac{116}{300} = 0.3867
    • Probability of defaulting and being single: P(DS)=41300=0.1367P(D \cap S) = \frac{41}{300} = 0.1367
  • Joint Probability Table: Summarizes joint probabilities.

  • Marginal Probabilities: Probabilities of individual events, calculated as sums of joint probabilities.

    • Probability that a customer is married: P(M)=143300=0.4766P(M) = \frac{143}{300} = 0.4766
  • Conditional Probability: The probability of event AA given that event BB has occurred.

    • Notation: P(AB)=P(AB)P(B)P(A | B) = \frac{P(A \cap B)}{P(B)}
    • P(BA)=P(AB)P(A)P(B | A) = \frac{P(A \cap B)}{P(A)}
    • Note: P(AB)P(BA)P(A | B) \neq P(B | A)
  • Application of Conditional Probabilities:

    • Probability of defaulting given the customer is married:
      • P(DM)=P(DM)P(M)=0.26330.4766=0.5524P(D | M) = \frac{P(D \cap M)}{P(M)} = \frac{0.2633}{0.4766} = 0.5524
    • Probability of defaulting given the customer is single:
      • P(DS)=P(DS)P(S)=0.13670.5234=0.2611P(D | S) = \frac{P(D \cap S)}{P(S)} = \frac{0.1367}{0.5234} = 0.2611
    • Conclusion: Default probability is higher if the customer is married.
  • Independent Events: Events AA and BB are independent if

    • P(AB)=P(A)P(A | B) = P(A)
    • P(BA)=P(B)P(B | A) = P(B)
    • Otherwise, AA and BB are dependent.
    • In the mortgage default case:
      • P(D)=0.40P(D) = 0.40
      • P(DM)=0.5524P(D | M) = 0.5524
      • P(DS)=0.2611P(D | S) = 0.2611
      • Since P(DM)P(D)P(D | M) \neq P(D), events DD and MM are dependent.
  • Multiplication Law:

    • P(AB)=P(AB)P(B)P(A \cap B) = P(A | B) \cdot P(B)
    • P(BA)=P(BA)P(A)P(B \cap A) = P(B | A) \cdot P(A)
    • For independent events:
      • P(AB)=P(A)P(B)P(A \cap B) = P(A) \cdot P(B)
      • P(BA)=P(B)P(A)P(B \cap A) = P(B) \cdot P(A)
      • Events A and B are dependent when P(AB)P(A)P(B)P(A \cap B) \neq P(A) \cdot P(B)
  • Bayes’ Theorem:

    • For two events, A<em>1A<em>1 and A</em>2A</em>2:
      • P(A<em>iB)=P(A</em>i)P(BA<em>i)P(A</em>1)P(BA<em>1)+P(A</em>2)P(BA2)P(A<em>i | B) = \frac{P(A</em>i) \cdot P(B | A<em>i)}{P(A</em>1) \cdot P(B | A<em>1) + P(A</em>2) \cdot P(B | A_2)} with i=1,2i = 1, 2
      • P(A<em>1)P(A<em>1) and P(A</em>2)P(A</em>2) are prior probabilities.
      • P(BA<em>1)P(B | A<em>1) and P(BA</em>2)P(B | A</em>2) are conditional probabilities.
      • Bayes’ theorem calculates revised (posterior) probabilities P(A<em>1B)P(A<em>1 | B) and P(A</em>2B)P(A</em>2 | B).

    5.4 Random Variables

  • Random variable: A numerical description of an experimental outcome.

  • Discrete random variable: May assume either a finite number of values or an infinite sequence of values.

    • Examples:
      • Flip a coin: Face of coin showing (1 if heads; 0 if tails)
      • Roll a die: Number of dots showing on top of die (1, 2, 3, 4, 5, 6)
      • Contact five customers: Number of customers who place an order (0, 1, 2, 3, 4, 5)
      • Operate a health care clinic for one day: Number of patients who arrive (0, 1, 2, 3, …)
      • Offer a customer the choice of two products: Product chosen by customer (0 if none; 1 if choose product A; 2 if choose product B)
  • Continuous random variable: May assume any numerical value in an interval or collection of intervals.

    • Examples:
      • Customer visits a web page: Time customer spends on web page in minutes ( x0x \geq 0 )
      • Fill a soft drink can (max capacity = 12.1 ounces): Number of ounces ( 0x12.10 \leq x \leq 12.1 )
      • Test a new chemical process: Temperature when the desired reaction takes place (min temperature = 150°F; max temperature = 212°F) ( 150x212150 \leq x \leq 212 )
      • Invest $10,000 in the stock market: Value of investment after one year ( x0x \geq 0 )

5.5 Discrete Probability Distributions

  • The probability distribution for a random variable describes the range and relative likelihood of possible values for a random variable.

  • For a discrete random variable xx, the probability distribution is defined by a probability mass function, denoted by f(x)f(x).

  • The probability mass function f(x)f(x) provides the probability for each value of the random variable xx.

  • Required conditions for a discrete probability function are: f(x)0f(x) \geq 0 and f(x)=1\sum f(x) = 1

  • Empirical Probability Distribution: A probability distribution generated from observations.

    • An empirical probability distribution is considered a custom discrete probability distribution if it is discrete and the possible values of the random variable have different probabilities.
    • The probabilities of a custom distribution can be generated using either the subjective method or the relative frequency method.
  • Custom Discrete Probability Distribution

    • Consider a random variable, xx, describing the number of payments made per year by 300 home mortgage customers:
      • 45 customers made quarterly payments (x=4x = 4), 180 made monthly payments (x=12x = 12), and 75 made two payments per month (x=24x = 24).
        • Probability distribution as a function:
          f(x)={0.15,x=4 0.60,x=12 0.25,x=24 0,otherwisef(x) = \begin{cases} 0.15, & x = 4 \ 0.60, & x = 12 \ 0.25, & x = 24 \ 0, & \text{otherwise} \end{cases}
  • Expected Value: The expected value, or mean, is a measure of the central location of a random variable, xx. It is calculated as the weighted average of the values of the random variable. The weights are the probabilities, f(x)f(x).

    • E(x)=μ=xf(x)E(x) = \mu = \sum x f(x)
  • Variance: The variance is a measure of variability in the values of a random variable, xx. It is calculated as the weighted average of the squared deviations of a random variable, xx, from its mean, μ\mu. The weights are the probabilities, f(x)f(x).

    • Var(x)=σ2=(xμ)2f(x)Var(x)=\sigma^2 = \sum(x-\mu)^2f(x)
  • Standard Deviation: The standard deviation σ\sigma is the positive square root of the variance.

    • σ=σ2=Var(x)\sigma = \sqrt{\sigma^2} = \sqrt{Var(x)}
  • Discrete Uniform Probability Distribution: The simplest example of a discrete probability distribution given by a formula.

    • The discrete uniform probability mass function can be written as:
      • f(x)=1nf(x) = \frac{1}{n}
      • Where nn is the number of values the random variable may assume.
  • Binomial Probability Distribution: The binomial probability function determines the probability of observing xx successes in nn independent events when there are only two possible outcomes.

    • The binomial probability mass function is described by the equation:
      • f(x)=(nx)px(1p)nxf(x) = \binom{n}{x} p^x (1-p)^{n-x}
      • Where pp is the probability of a success in a single event, and 1p1-p the probability of a failure in a single event.
  • Poisson Probability Distribution: A Poisson Probability Distribution describes the number of occurrences, xx, over a specified interval of time or space, if the following two properties are met:

    • 11. The probability of an occurrence is the same for any two intervals (time or space) of equal length.
    • 22. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

5.6 Continuous Probability Distributions

  • Probability mass function provides the probability that the random variable assumes a particular value for Discrete random variables
  • Probability density function is the counterpart of the probability mass function for Continuous random variables.
  • Uniform Probability Distribution: A continuous random variable, xx, is described by a uniform probability distribution whenever the probability is proportional to the interval’s length.
    • The uniform probability density function is
      f(x)={1bafor axb 0elsewheref(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \leq x \leq b \ 0 & \text{elsewhere} \end{cases}
  • Triangular Probability Distribution: A continuous random variable, xx, can be described by a triangular probability distribution when only the following subjective information is available:
    • the minimum possible value, aa,
    • the maximum possible value, bb, and
    • the most likely value (or mode) of the distribution, mm.
  • Normal Probability Distribution: The normal probability distribution assumes the characteristic bell-shaped curve and is the most common probability continuous distribution.
    • The normal probability density function is entirely described by two parameters: the mean, μ\mu, and the standard deviation, σ\sigma
      • f(x)=12πσ2e12(xμσ)2f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} (\frac{x-\mu}{\sigma})^2}
    • Where π3.14159\pi \approx 3.14159 and e2.71828e \approx 2.71828
  • Exponential Probability Distribution: The exponential probability distribution is used for a random variable, xx, that describes the length of an interval between occurrences.
    • The exponential probability density function is
      • f(x)=1μexμfor x0f(x) = \frac{1}{\mu} e^{-\frac{x}{\mu}} \quad \text{for } x \geq 0