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 : Numerical outcome of random process
Discrete random variable: Countable set of possible values (integers typically)
Examples:
- Number of heads in 3 coin flips:
- Sum of two dice:
- Number of free throws made in 10 attempts
Probability Mass Function (PMF)
Definition: for each possible value
Requirements:
- for all
- (probabilities sum to 1)
Example: Number of heads in 3 fair coin flips
| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
| 1/8 | 3/8 | 3/8 | 1/8 |
Check: ✓
Expected Value (Mean)
Interpretation: Long-run average value
Example:
Intuition: On average, 1.5 heads in 3 flips
Variance and Standard Deviation
Example: For 3 coin flips ()
Linear Combinations of Random Variables
If :
If (independent):
If (independent): (subtract in means, add in variances!)
Common Mistakes
- Forgetting variance is additive even for differences (, not minus)
- Using 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.
⚠️ Common Mistakes: Discrete Random Variables
Avoid these 3 frequent errors
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