Probability Distributions: Discrete Distributions

This lesson introduces you to discrete probability distributions, which describe the probabilities of different outcomes in a discrete random variable. You'll learn about key discrete distributions like the binomial and Poisson distributions and how to apply them to solve real-world problems.

Learning Objectives

  • Define and differentiate between discrete and continuous random variables.
  • Understand the concept of probability mass function (PMF).
  • Describe and apply the binomial distribution to solve problems.
  • Describe and apply the Poisson distribution to solve problems.

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Lesson Content

Introduction to Discrete Random Variables

A random variable is a variable whose value is a numerical outcome of a random phenomenon. A discrete random variable is a variable that can only take on a finite number of values, or a countably infinite number of values (like integers). Think of it as 'countable' – you can list out all the possible outcomes. Examples include the number of heads when flipping a coin a few times, the number of cars passing a certain point on the road in an hour, or the number of defective items in a sample. Conversely, a continuous random variable can take on any value within a range (e.g., height, weight, temperature).

Probability Mass Function (PMF)

The Probability Mass Function (PMF), often denoted as P(X = x), defines the probability that a discrete random variable, X, takes on a specific value, x. For each possible value of x, the PMF assigns a probability. Key properties of a PMF are:

  • The probability for each value is between 0 and 1 (inclusive).
  • The sum of probabilities for all possible values must equal 1.

Example: Imagine flipping a fair coin twice. Let X be the number of heads. The possible values for X are 0, 1, and 2. The PMF would be:

  • P(X = 0) = 1/4 (TT)
  • P(X = 1) = 2/4 = 1/2 (HT, TH)
  • P(X = 2) = 1/4 (HH)

Binomial Distribution

The Binomial Distribution models the number of successes in a fixed number of independent trials, where each trial has only two possible outcomes: success or failure. It's characterized by:

  • n: The number of trials (fixed).
  • p: The probability of success on a single trial (constant).

The formula for the binomial probability mass function is:

P(X = k) = (nCk) * p^k * (1-p)^(n-k)

Where:

  • X is the number of successes.
  • k is the number of successes we're interested in.
  • nCk is the binomial coefficient, read as "n choose k," which represents the number of ways to choose k successes from n trials. You can calculate it as n! / (k! * (n-k)!) where '!' denotes factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1).
  • p^k is the probability of getting k successes.
  • (1-p)^(n-k) is the probability of getting (n-k) failures.

Example: Suppose you flip a fair coin 5 times (n=5). What's the probability of getting exactly 3 heads (k=3)? p = 0.5 (probability of heads).

P(X = 3) = (5C3) * 0.5^3 * (0.5)^2 = (10) * 0.125 * 0.25 = 0.3125

Poisson Distribution

The Poisson Distribution models the number of events that occur in a fixed interval of time or space, given that these events happen independently and at a constant average rate. It's often used for rare events. It's characterized by:

  • λ (lambda): The average rate of events per interval (e.g., events per hour, calls per minute).

The formula for the Poisson probability mass function is:

P(X = k) = (e^(-λ) * λ^k) / k!

Where:

  • X is the number of events.
  • k is the number of events we're interested in.
  • λ is the average rate of events.
  • e is Euler's number (approximately 2.71828).
  • k! is the factorial of k.

Example: A call center receives an average of 4 calls per hour (λ=4). What's the probability of receiving exactly 2 calls in an hour (k=2)?

P(X = 2) = (e^(-4) * 4^2) / 2! ≈ (0.0183 * 16) / 2 ≈ 0.1464

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