Discrete Probability Distributions

• This document covers statistical analysis with software applications, focusing on discrete probability distributions, random variables, expected value, and variance.
• Rocelle Ann G. Terco is the facilitator for course AE 09 (Lec).

Learning Objectives

• Illustrate random variables, distinguishing between discrete and continuous types.
• Determine possible values for random variables.
• Illustrate probability distributions.
• Construct probability mass functions.
• Compute probabilities.

Flow of Session

• Review fundamental concepts.
• Receive an overview of the topic.
• Study examples of random variables using probability distributions.
• Learn about expected value and variance.
• Complete a checkpoint exercise.

Discrete Probability Distribution

• Chapter V focuses on discrete probability distributions.
• Random variables, which have properties and types, may be described by a probability distribution.
• Types of probability distribution include discrete and continuous, with discrete distributions having probability mass functions.

Random Variables

• Random variables represent variables whose possible values are numerical outcomes of a random experiment.

Types of Random Variables

• Discrete variables consist of countable, specific values and include whole numbers.
• Continuous variables are measurable, can take any value within a range, and include real numbers.

Examples of Random Variables

• 'A' represents the sum of numbers when a pair of dice is tossed.
• 'B' represents the distance leaped (in meters) by a long-jumper.
• 'W' represents the length of time (in minutes) a scheduled airplane flight is delayed.
• 'X' represents the number of correct answers on a 10-item True-False test.
• A and X are classified as discrete random variables.
• B and W are classified as continuous random variables.

Probability Distribution Function

• A probability distribution demonstrates the relative likelihood of each outcome in an experiment.
• Consider a football team playing three consecutive games, where 'W' is a win and 'L' is a loss.
• The sample space is the set of all possible outcomes.
• The sample space is {WWW, WWL, WLW, LWW, WLL, LWL, LLW, LLL}.

Sample Point Probabilities

• WWW (3 wins)
• WWL, WLW, LWW (2 wins)
• WLL, LWL, LLW (1 win)
• LLL (0 wins)

Probability Mass Function

• A probability mass function (PMF) assigns a probability to each sample point.
• PMFs can be expressed in tabular, graphical, or formula form.
• The probability of each value of the discrete random variable is between 0 and 1, inclusive.
• The sum of all probabilities equals 1.

Example: Dice Game

• A pair of dice thrown, and the sum of the values determines winning.
• A win occurs with a sum of 3 to 9.

Dice Game Sample Space

• The sample space includes all possible sums from 2 to 12 when throwing two dice.
• 2nd Die (Columns): 1, 2, 3, 4, 5, 6
• 1st Die (Rows): 1, 2, 3, 4, 5, 6
• Sum of the values of the 2 die face up: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
• The number of points within the probability mass function is listed from 1 to 6 then down to 1

Probability Calculation

• The probability of getting a sum of 3 to 9 in the dice game is P(3 ≤ Y ≤ 9).
• P(3 ≤ Y ≤ 9) = 2/36 + 3/36 + 4/36 + 5/36 + 6/36 + 5/36 + 4/36 = 29/36.
• The probability of winning the dice game is 29/36, or approximately 80.56%.

Checkpoint 1: Absolute Value of Dice Differences

• Objective: Construct the probability mass function (PMF) of Y and compute related probabilities in an experiment involving rolling a pair of dice.
• Y represents the random variable of the absolute value of the difference of the numbers that come up.
• Find the probability of randomly getting an absolute value that is (a) two, (b) at most three, and (c) between two and five.
• Six possible values of Y are: 0, 1, 2, 3, 4, and 5 – each from 36 sample points.

Expected Value (Mean) of a Discrete Random Variable

• The mean (μ) of a random variable, also known as the expected value (E), represents the anticipated average outcome over many trials.
• For a discrete random variable X, the expected value is calculated as E(X) = μ = Σ [xi * P(xi)], where xi is each value of X.
• To calculate the mean, multiply each value of X by its corresponding probability and add the products.

Variance and Standard Deviation

• Variance measures the spread of different values about the mean.
• Alternative formula: σ² = μ(Χ2) – (μ(Χ))^2
• Standard deviation is the square root of the variance (σ = √σ²).

Example: Reunion Game Fair Price

• A high school reunion committee organizes a game with 150 balls in a box to raise funds.
• The ball values includes ₱500, ₱1000, ₱5000 and ₱0
• 10 balls win ₱500, 5 win ₱1000, 1 wins ₱5000, and 134 win nothing.

Solution

• Define A is the random variable that represents the amount a player can win.
• Then, A can take on the values 500, 1000, 5000, and 0.
• These values of A, represent the random variable multiplied by its relevant propbability with the corresponding result
• E(A) = 0*(67/75) + 500*(1/15) + 1000*(1/30) + 5000*(1/150) = 100
• A fair price to play this game is Php100.

Example 2: Car Insurance

• A car insurance company pays out Php500,000 for a stolen or destroyed car.
• The insurance policy costs Php24,000.
• The company pays out 0.002 percent of the time.

Solution

• Define X as the owner's net gain with possible values
• The payout is 476,000 and the loss is -24,000
• The value of X= 476,000 with 0.002 probability
• The value of X= −24,000 with 0.998 probability (1-0.002)
• Calculate the expect value for the insurance company:
• E(X) = (476,000)(0.002) + (-24,000)(0.998) = −23,000
• The policy favors the company, not the owner.

Example 4: Absolute Difference of Dice

• Consider random variable Y, which is the absolute value of the difference between the numbers when rolling a pair of dice.
• The expected value of Y is approximately 1.94.
• The variance of Y is approximately 2.05.
• Standard deviation of Y = √2.05 ≈ 1.43.

Binomial Distribution

• Describes a distribution with only two possible outcomes.
• p = probability of success
• q = probability of failure
• n = number of trials
• Formula P(x) = C(n, r) *p^x * q^(n-x), where C(n, r) = n! / (r!(n-r)!).

Poisson Distribution

• Models how many times an event is likely to occur over a specified period.
• Lambda is equal to the expected value (EV) of x when that is also equal to its variance.
• e is Euler's number (e = 2.71828...)
• x is the number of occurrences
• x! is the factorial of x
• Formula P(X) = (λ^x * e^-λ) / X!