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Discrete Random Variables

Define discrete random variables, calculate expected value, variance, and standard deviation.

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📊 Discrete Random Variables

Definition and Notation

Random variable XX: Numerical outcome of random process

Discrete random variable: Countable set of possible values (integers typically)

Examples:

  • Number of heads in 3 coin flips: X∈{0,1,2,3}X \in \{0, 1, 2, 3\}
  • Sum of two dice: X∈{2,3,...,12}X \in \{2, 3, ..., 12\}
  • Number of free throws made in 10 attempts

Probability Mass Function (PMF)

Definition: P(X=x)P(X = x) for each possible value xx

Requirements:

  • 0≤P(X=x)≤10 ≤ P(X = x) ≤ 1 for all xx
  • ∑P(X=x)=1\sum P(X = x) = 1 (probabilities sum to 1)

Example: Number of heads in 3 fair coin flips

xx0123
P(X=x)P(X=x)1/83/83/81/8

Check: 1/8+3/8+3/8+1/8=11/8 + 3/8 + 3/8 + 1/8 = 1 ✓

Expected Value (Mean)

E(X)=μX=∑x⋅P(X=x)E(X) = \mu_X = \sum x · P(X = x)

Interpretation: Long-run average value

Example: E(X)=0(1/8)+1(3/8)+2(3/8)+3(1/8)=0+3/8+6/8+3/8=12/8=1.5E(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 0 + 3/8 + 6/8 + 3/8 = 12/8 = 1.5

Intuition: On average, 1.5 heads in 3 flips

Variance and Standard Deviation

Var(X)=σX2=∑(x−μ)2⋅P(X=x)=E(X2)−[E(X)]2\text{Var}(X) = \sigma_X^2 = \sum (x - \mu)^2 · P(X = x) = E(X^2) - [E(X)]^2

σX=Var(X)\sigma_X = \sqrt{\text{Var}(X)}

Example: For 3 coin flips (μ=1.5\mu = 1.5)

Var(X)=(0−1.5)2(1/8)+(1−1.5)2(3/8)+(2−1.5)2(3/8)+(3−1.5)2(1/8)\text{Var}(X) = (0-1.5)^2(1/8) + (1-1.5)^2(3/8) + (2-1.5)^2(3/8) + (3-1.5)^2(1/8) =2.25(1/8)+0.25(3/8)+0.25(3/8)+2.25(1/8)=0.75= 2.25(1/8) + 0.25(3/8) + 0.25(3/8) + 2.25(1/8) = 0.75 σX=0.75≈0.866\sigma_X = \sqrt{0.75} ≈ 0.866

Linear Combinations of Random Variables

If W=aX+bW = aX + b: E(W)=aE(X)+bE(W) = aE(X) + b Var(W)=a2Var(X)\text{Var}(W) = a^2 \text{Var}(X)

If Y=X1+X2Y = X_1 + X_2 (independent): E(Y)=E(X1)+E(X2)E(Y) = E(X_1) + E(X_2) Var(Y)=Var(X1)+Var(X2)\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_2)

If Y=X1−X2Y = X_1 - X_2 (independent): E(Y)=E(X1)−E(X2)E(Y) = E(X_1) - E(X_2) Var(Y)=Var(X1)+Var(X2)\text{Var}(Y) = \text{Var}(X_1) + \text{Var}(X_2) (subtract in means, add in variances!)

Common Mistakes

  • Forgetting variance is additive even for differences (+Var(X2)+ \text{Var}(X_2), not minus)
  • Using Var(X1+X2)=Var(X1)+Var(X2)\text{Var}(X_1 + X_2) = \text{Var}(X_1) + \text{Var}(X_2) when variables are dependent
  • Confusing PMF with CDF

Decision Rule

Modify random variable? → Apply linearity rules Independent sums? → Add means and variances

AP Exam Tip

"Find the probability distribution of X" means: create table with all possible values and their probabilities. Always verify probabilities sum to 1.

📚 Practice Problems

1Problem 1easy

❓ Question:

Let X = number of heads in 2 coin flips. List the probability distribution of X and verify it sums to 1.

💡 Show Solution

Outcomes: TT (0 heads), TH/HT (1 head), HH (2 heads). X | 0 | 1 | 2 P(X) | 0.25| 0.5 | 0.25 Sum = 0.25 + 0.5 + 0.25 = 1.0. ✓ This is a valid discrete distribution. The probabilities are non-negative and sum to 1.

2Problem 2medium

❓ Question:

A discrete random variable has distribution: P(X=1)=0.1, P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.2. Find P(X ≤ 2) and P(X > 2).

💡 Show Solution

P(X ≤ 2) = P(X=1) + P(X=2) = 0.1 + 0.3 = 0.4. P(X > 2) = P(X=3) + P(X=4) = 0.4 + 0.2 = 0.6. Check: P(X ≤ 2) + P(X > 2) = 0.4 + 0.6 = 1.0. ✓ The cumulative probability P(X ≤ 2) represents 40% of the distribution.

3Problem 3hard

❓ Question:

A test score X has distribution: P(X=60)=0.1, P(X=70)=0.25, P(X=80)=0.4, P(X=90)=0.2, P(X=100)=0.05. Find E(X) and Var(X).

💡 Show Solution

E(X) = 60(0.1) + 70(0.25) + 80(0.4) + 90(0.2) + 100(0.05) = 6 + 17.5 + 32 + 18 + 5 = 78.5. E(X²) = 3600(0.1) + 4900(0.25) + 6400(0.4) + 8100(0.2) + 10000(0.05) = 360 + 1225 + 2560 + 1620 + 500 = 6265. Var(X) = E(X²) - [E(X)]² = 6265 - (78.5)² = 6265 - 6162.25 = 102.75. The mean test score is 78.5, with variance measuring typical squared deviation from this mean.

Explain using:

⚠️ Common Mistakes: Discrete Random Variables

Avoid these 3 frequent errors

📌 Related Topics in Unit 4: Probability, Random Variables, and Probability Distributions

❓ Frequently Asked Questions

What is Discrete Random Variables?▾
Define discrete random variables, calculate expected value, variance, and standard deviation.
How can I study Discrete Random Variables effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Discrete Random Variables?▾
Discrete Random Variables is part of the AP Statistics course on Study Mondo, specifically in the Unit 4: Probability, Random Variables, and Probability Distributions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Discrete Random Variables?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.