Lesson 4: Probability Distributions
Module 3: Probability Basics • Lesson 4 of 4 • ~30-35 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define random variables and distinguish between discrete and continuous
- Create and interpret probability distributions
- Verify that a distribution is valid
- Calculate expected value (mean) of a probability distribution
- Calculate variance and standard deviation of a probability distribution
Random Variables
Random Variable
A random variable is a variable whose value is determined by the outcome of a random experiment. We usually denote random variables with capital letters like X, Y, Z.
Examples of Random Variables:
- X = number of heads when flipping 3 coins
- Y = sum when rolling two dice
- Z = number of customers who enter a store in an hour
- W = height of a randomly selected student
Types of Random Variables
Discrete Random Variable
A discrete random variable can only take on specific, countable values (usually whole numbers).
Examples: Number of heads (0, 1, 2, 3), dice sum (2, 3, ..., 12), number of customers (0, 1, 2, ...)
Continuous Random Variable
A continuous random variable can take on any value within a range (infinite possible values).
Examples: Height (5.1", 5.7", 5.93", ...), time (3.5 sec, 3.52 sec, ...), temperature
In this lesson, we'll focus on discrete random variables and their probability distributions.
Probability Distributions
Probability Distribution
A probability distribution lists all possible values of a random variable along with their probabilities.
It shows the complete picture of what outcomes are possible and how likely each one is.
Example 1: Flipping Two Coins
Let X = number of heads when flipping two fair coins
Sample space: {HH, HT, TH, TT}
Possible values of X: 0, 1, 2
| X (# heads) | Outcomes | P(X) |
|---|---|---|
| 0 | TT | 1/4 = 0.25 |
| 1 | HT, TH | 2/4 = 0.50 |
| 2 | HH | 1/4 = 0.25 |
| Total | 1.00 | |
This table is the probability distribution of X!
Requirements for a Valid Probability Distribution
For a valid probability distribution:
- All probabilities must be between 0 and 1: 0 ≤ P(X = x) ≤ 1
- All probabilities must sum to 1: ΣP(X = x) = 1
Example 2: Is This a Valid Distribution?
| X | P(X) |
|---|---|
| 1 | 0.2 |
| 2 | 0.3 |
| 3 | 0.5 |
| Sum: | 1.0 |
Answer: YES!
- All probabilities between 0 and 1
- Sum = 0.2 + 0.3 + 0.5 = 1.0
Example 3: Is This Valid?
| X | P(X) |
|---|---|
| 1 | 0.3 |
| 2 | 0.4 |
| 3 | 0.5 |
| Sum: | 1.2 |
Answer: NO!
- All probabilities between 0 and 1
- Sum = 0.3 + 0.4 + 0.5 = 1.2 ≠ 1.0
This violates the requirement that probabilities sum to exactly 1!
Expected Value (Mean)
Expected Value
The expected value (or mean) of a random variable X is the long-run average value if the experiment were repeated many times.
It's denoted as E(X), μ, or μ_X.
Expected Value Formula
E(X) = μ = Σ[x · P(X = x)]
In words: Multiply each value by its probability, then add them all up
Understanding Expected Value
The expected value is like a weighted average:
- Values with higher probabilities "pull" the average toward them
- It's the theoretical mean if you repeated the experiment infinitely many times
- Important: E(X) doesn't have to be a possible value of X!
Example 4: Expected Number of Heads
Using our coin-flipping distribution from earlier:
| X | P(X) | x · P(X) |
|---|---|---|
| 0 | 0.25 | 0 × 0.25 = 0.00 |
| 1 | 0.50 | 1 × 0.50 = 0.50 |
| 2 | 0.25 | 2 × 0.25 = 0.50 |
| E(X) = Sum: | 1.00 | |
Answer: E(X) = 1.00 head
On average, we expect to get 1 head when flipping 2 coins (makes sense - it's the middle value!)
Example 5: Expected Winnings (Game)
Game: Roll a die. Win $10 for rolling a 6, lose $2 otherwise.
Let X = winnings
| X (winnings) | P(X) | x · P(X) |
|---|---|---|
| −$2 | 5/6 ≈ 0.833 | −2 × (5/6) = −10/6 |
| $10 | 1/6 ≈ 0.167 | 10 × (1/6) = 10/6 |
| E(X) = Sum: | 0/6 = $0.00 | |
Answer: E(X) = $0
Interpretation: This is a "fair game" - on average, you neither win nor lose money in the long run!
Example 6: Insurance Expected Value
Scenario: An insurance company sells a $100,000 life insurance policy for $500. Actuaries estimate a 0.002 probability the policyholder will die this year.
Let X = company's profit
| Outcome | Profit X | P(X) | x · P(X) |
|---|---|---|---|
| No claim | $500 | 0.998 | 500 × 0.998 = 499 |
| Claim filed | 500 − 100,000 = −$99,500 | 0.002 | −99,500 × 0.002 = −199 |
| E(X) = Sum: | $300 | ||
Answer: E(X) = $300 profit per policy
Business decision: On average, the company expects to make $300 per policy sold!
Variance and Standard Deviation
Variance and Standard Deviation
The variance (σ² or Var(X)) measures the spread or variability of the distribution around the mean.
The standard deviation (σ or SD(X)) is the square root of variance, giving spread in the original units.
Variance Formula
Var(X) = σ² = Σ[(x − μ)² · P(X = x)]
In words: For each value, find (value − mean)², multiply by probability, sum them up
Standard Deviation Formula
SD(X) = σ = √Var(X)
Connection to Module 2
These formulas are similar to variance and SD from Module 2!
- Module 2: Calculated from actual data values
- Module 3: Calculated from probability distribution (theoretical)
- Both measure spread around the mean!
Example 7: Variance of Coin Flips
Using our coin-flipping distribution (recall E(X) = 1):
| X | P(X) | (x − μ) | (x − μ)² | (x − μ)² · P(X) |
|---|---|---|---|---|
| 0 | 0.25 | 0 − 1 = −1 | 1 | 1 × 0.25 = 0.25 |
| 1 | 0.50 | 1 − 1 = 0 | 0 | 0 × 0.50 = 0.00 |
| 2 | 0.25 | 2 − 1 = 1 | 1 | 1 × 0.25 = 0.25 |
| Var(X) = Sum: | 0.50 | |||
Variance: σ² = 0.50
Standard Deviation: σ = √0.50 ≈ 0.71
Check Your Understanding
Practice Question 1
Is X = "height of a student" discrete or continuous?
Answer: Continuous
Height can take on any value within a range (5'7.3", 5'7.31", etc.), not just specific countable values.
Practice Question 2
Is this a valid probability distribution?
| X | P(X) |
|---|---|
| 0 | 0.1 |
| 1 | 0.4 |
| 2 | 0.3 |
| 3 | 0.2 |
Answer: YES, it's valid
All probabilities between 0 and 1
Sum = 0.1 + 0.4 + 0.3 + 0.2 = 1.0
Practice Question 3
Calculate E(X) for this distribution:
| X | P(X) |
|---|---|
| 1 | 0.2 |
| 2 | 0.5 |
| 3 | 0.3 |
Solution:
E(X) = Σ[x · P(X)]
= (1 × 0.2) + (2 × 0.5) + (3 × 0.3)
= 0.2 + 1.0 + 0.9
= 2.1
Practice Question 4
True or False: The expected value must be one of the possible values of X.
Answer: FALSE
Explanation:
Expected value is a weighted average and doesn't have to be a possible value!
Example: When flipping 2 coins, E(X) = 1.0 head, but you can't get exactly 1.0 heads - only 0, 1, or 2!
Practice Question 5
If a game has E(X) = −$2, what does this mean?
Answer:
On average, you expect to LOSE $2 per game in the long run.
This is not a favorable game for the player! The negative expected value means the game is biased against you.
Key Takeaways
- Random variable (X): Variable whose value depends on chance
- Discrete: Countable values; Continuous: Any value in a range
- Probability distribution: Lists all values of X with their probabilities
- Valid distribution: All P(X) between 0 and 1, sum = 1
- Expected value E(X) = μ: Long-run average, weighted by probabilities
- Variance σ²: Measures spread around the mean
- Standard deviation σ: Square root of variance, same units as X
- Connection to Module 2: Same concepts (mean, variance, SD), but calculated from probabilities instead of data!
Congratulations!
You've completed all four lessons in Module 3: Probability Basics! You now understand:
- Basic probability and probability rules
- Counting methods (permutations and combinations)
- Conditional probability and independence
- Probability distributions and expected value
Next steps: Practice your skills with comprehensive problems, then take the Module 3 Quiz!