Discrete Random Variables

🎲 Discrete Random Variables

Part 1 of 7 — Probability Distributions

What Is a Random Variable?

A random variable assigns a numerical value to each outcome of a random process.

🔑 Key Idea: A discrete random variable has a probability distribution that lists every possible value and its probability.

Probability Distribution Table

Requirements:

• Every probability is between 0 and 1: $0 \leq P(X = x) \leq 1$
• All probabilities sum to 1: $\sum P(X = x) = 1$

Reading the Table

From the table above:

• $P(X = 2) = 0.4$
• $P(X \geq 2) = P(X=2) + P(X=3) = 0.4 + 0.2 = 0.6$
• $P(X < 2) = P(X=0) + P(X=1) = 0.1 + 0.3 = 0.4$

Concept Check 🎯

Probability Distribution Practice 🧮

Given: $P(X=1) = 0.25$, $P(X=2) = 0.35$, $P(X=3) = 0.30$, $P(X=4) = 0.10$

1) $P(X \leq 2)$?

2) $P(X > 2)$?

3) Do the probabilities sum to 1? (yes or no)

🎯 Expected Value (Mean of a Random Variable)

Part 2 of 7 — Expected Value

The Mean of a Discrete Random Variable

$\mu_X = E(X) = \sum x_i \cdot P(x_i)$

The expected value is the long-run average — if you repeated the random process many times, the average outcome would approach $E(X)$.

Worked Example

$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7$

⚠️ The expected value does NOT have to be a possible outcome. $X$ can’t actually equal 1.7, but 1.7 is the long-run average.

Interpretation on the AP Exam

“If we were to repeat this random process many, many times, the average value of $X$ would be approximately 1.7.”

Concept Check 🎯

Expected Value Calculation 🧮

A game costs $5 to play. You win $20 with probability 0.2, win $5 with probability 0.3, and win $0 with probability 0.5.

Let $X$ = net gain (winnings minus cost).

1) $X$ values: $15, $0, and $\underline{\quad}$ (third net gain value)

2) $E(X) = ?$ (expected net gain)

3) Is this game favorable for the player? (yes or no)

📊 Variance & Standard Deviation of a Random Variable

Part 3 of 7 — Spread of a Distribution

Variance

$\text{Var}(X) = \sigma_X^2 = \sum (x_i - \mu_X)^2 \cdot P(x_i)$

Standard Deviation

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

Worked Example

Using our distribution: $\mu_X = 1.7$

$\sigma^2 = 0.289 + 0.147 + 0.036 + 0.338 = 0.81$ $\sigma = \sqrt{0.81} = 0.9$

🔑 Standard deviation measures the typical distance of outcomes from the mean.

Concept Check 🎯

Variance Drill 🧮

A random variable $X$ has: $\mu = 3$, and outcomes $X = 1, 3, 5$ each with probability $1/3$.

1) $(1 - 3)^2 \cdot (1/3) = ?$ (first term of variance calculation)

2) $(3 - 3)^2 \cdot (1/3) = ?$

3) $\sigma^2 = ?$ (total variance, as a fraction like a/b)

⚖️ Cumulative Distributions and Simulation

Part 4 of 7 — Building and Estimating Distributions

Building a Distribution with Probability Rules

A bag holds 3 red and 2 blue marbles. Draw 2 without replacement and let $X$ = number of red marbles.

• $P(X = 0) = \frac{2}{5} \cdot \frac{1}{4} = 0.10$
• $P(X = 2) = \frac{3}{5} \cdot \frac{2}{4} = 0.30$
• $P(X = 1) = \frac{3}{5} \cdot \frac{2}{4} + \frac{2}{5} \cdot \frac{3}{4} = 0.60$ (red then blue, or blue then red)

Cumulative Probability Distribution

A cumulative probability distribution lists $P(X \leq x)$ for each value $x$:

• Each cumulative value adds the next probability.
• The last cumulative value is always 1.
• Recover an exact probability by subtracting neighbors: $P(X = 1) = 0.70 - 0.10 = 0.60$.

Concept Check 🎯

Estimating a Distribution by Simulation

When the rules are hard to apply, simulate.

Setting: Each cereal box holds one of 3 prizes, equally likely. Let $X$ = number of different prizes in 4 boxes ($X$ = 1, 2 or 3).

Simulation with random digits:

1. Let digits 1, 2, 3 stand for the three prizes; skip 0 and 4–9.
2. Read four usable digits (one per box) and record how many different prizes appear.
3. Repeat many times.

Results of 50 repetitions:

The exact probabilities are 0.037, 0.519 and 0.444. With more repetitions, the estimates tend to get closer to them.

Simulation Practice 🧮

A different simulation of the prize setting gives these counts in 80 repetitions: $X = 1$: 4, $X = 2$: 40, $X = 3$: 36.

1) Estimated $P(X = 2)$? (2 decimals)

2) Estimated $P(X \leq 2)$? (2 decimals)

3) Estimated $P(X \geq 2)$? (2 decimals)

🔄 Interpreting Expected Value and Standard Deviation

Part 5 of 7 — Using $\mu_X$ and $\sigma_X$ in Context

Interpretation Templates

Both are parameters: single fixed numbers that describe the distribution, given in the units of $X$.

Worked Example: A Raffle

A school sells 500 raffle tickets at $2 each. One ticket wins $300 and four tickets win $50. Let $X$ = net gain for one ticket.

$E(X) = 298(0.002) + 48(0.008) + (-2)(0.990) = 0.596 + 0.384 - 1.98 = -1.00$

Interpretation: Over many tickets, buyers lose about $1.00 per ticket on average. (The school raises about $1 per ticket.)

Concept Check 🎯

Using $\sigma_X$ to Compare Options

Two options can have the same mean but very different risk.

On average both routes take 30 minutes, but Route B's times typically vary by about 9 minutes from that mean, so Route B is much less predictable. If a package must arrive within 35 minutes, Route A is the safer choice.

🔑 Use the mean to compare typical outcomes and the standard deviation to compare consistency or risk.

Decisions with $\mu$ and $\sigma$ 🧭

Mean and SD in Context 🧮

The number of defective items in a box, $X$, has $P(X=0) = 0.7$, $P(X=1) = 0.2$, $P(X=2) = 0.1$.

1) $E(X)$?

2) $\sigma_X$? (round to 2 decimals)

3) Over 100 boxes, about how many defective items in total would you expect?

🏆 Problem-Solving Workshop

Part 6 of 7 — AP-Style Problems

Strategy for Random Variable Problems

1. Define the random variable in context and list its possible values
2. Check that the probabilities are between 0 and 1 and sum to 1
3. Calculate $E(X)$ using $\sum x \cdot P(x)$
4. Calculate $\sigma_X$ using $\sqrt{\sum (x - \mu_X)^2 \cdot P(x)}$
5. Interpret in context for full AP credit

AP Exam Tip

When asked to interpret expected value: “If [process] were repeated many times, the average [variable] would be approximately [value].”

When asked to interpret standard deviation: “The [variable] typically varies by about [value] from the mean of [mean].”

Concept Check 🎯

Insurance Problem 🧮

An insurance company charges $300/year. Claims: $0 (prob 0.9), $1000 (prob 0.08), $5000 (prob 0.02).

1) Expected claim per customer?

2) Expected profit per customer?

3) Standard deviation of claims? (round to nearest dollar)

📝 Review & Applications

Part 7 of 7 — Comprehensive Review

Key Formulas Summary

Common Mistakes on the AP Exam

• Averaging the values without weighting them by their probabilities
• Forgetting the square root when going from variance to standard deviation
• Thinking $E(X)$ has to be a possible value
• Confusing “expected value” with “most likely value”
• Interpreting without context or units

Concept Check 🎯

Final Challenge 🧮

$X$ has $P(X=0) = 0.3$, $P(X=1) = 0.4$, $P(X=4) = 0.3$.

1) $E(X) = ?$

2) $\text{Var}(X) = ?$ (2 decimals)

3) $\sigma_X = ?$ (round to 2 decimals)