Basic Probability
Probability measures how likely an event is to occur. It is always a number between 0 and 1 — where 0 means impossible and 1 means certain.
Experiment: A process with uncertain outcomes (rolling a die, flipping a coin).
Sample Space (S): The set of all possible outcomes.
Event (A): Any subset of the sample space.
P(A): The probability that event A occurs.
When exact probabilities are hard to calculate, we can simulate the experiment many times and use the relative frequency as an estimate. The Law of Large Numbers says that as the number of trials increases, the simulated probability gets closer to the true probability.
Probability Rules
Subtract the overlap to avoid counting it twice.
Mutually exclusive: A and B cannot both happen. If A occurs, B cannot. P(A ∩ B) = 0.
Independent: Knowing A occurred tells you nothing about B. P(A ∩ B) = P(A) · P(B). These are completely different concepts — mutually exclusive events are actually dependent (if A happens, you know B didn't).
P(drawing a heart) = 13/52 = 0.25 P(drawing a face card) = 12/52 ≈ 0.231
P(heart AND face card) = 3/52 ≈ 0.058 (Jack, Queen, King of hearts)
P(heart OR face card) = 0.25 + 0.231 − 0.058 = 0.423
Conditional Probability & Independence
Restricts the sample space to only the outcomes in B.
A and B are independent if any of these equivalent conditions hold:
\(P(A \mid B) = P(A)\) — knowing B happened doesn't change probability of A
\(P(B \mid A) = P(B)\) — knowing A happened doesn't change probability of B
\(P(A \cap B) = P(A) \cdot P(B)\) — the multiplication rule holds
Two-Way Tables and Conditional Probability
Two-way tables are a powerful tool for calculating conditional probabilities on the AP exam.
| Plays Sport | Does Not Play | Total | |
|---|---|---|---|
| Grade 11 | 45 | 30 | 75 |
| Grade 12 | 35 | 40 | 75 |
| Total | 80 | 70 | 150 |
Using the table above:
P(plays sport) = 80/150 = 0.533
P(Grade 11 | plays sport) = 45/80 = 0.5625 → restrict to the "plays sport" column
P(plays sport | Grade 12) = 35/75 = 0.467 → restrict to the "Grade 12" row
Are grade and sport independent? P(sport) = 0.533. P(sport | Grade 11) = 45/75 = 0.60 ≠ 0.533. So not independent — grade and sport participation are associated.
Random Variables & Probability Distributions
A random variable X assigns a numerical value to each outcome of a random process.
Discrete RV: Takes a countable number of values (0, 1, 2, 3, …). Think counts.
Continuous RV: Takes any value in an interval. Think measurements (height, time, weight).
A table is a valid probability distribution if and only if:
1. Every probability is between 0 and 1: \(0 \leq P(X = x) \leq 1\)
2. All probabilities sum to exactly 1: \(\sum P(X = x) = 1\)
Expected Value & Standard Deviation of a Random Variable
Also called the mean of the distribution.
A game: roll a die. Win $10 if you roll a 6, lose $2 otherwise.
| Outcome | x (winnings) | P(X = x) | x · P(x) |
|---|---|---|---|
| Roll a 6 | +$10 | 1/6 | +10/6 ≈ 1.667 |
| Roll 1–5 | −$2 | 5/6 | −10/6 ≈ −1.667 |
\(\mu_X = \frac{10}{6} + \frac{-10}{6} = \mathbf{0}\)
Expected value = $0. This is a fair game — on average, neither player gains or loses money in the long run.
Rules for Combining Random Variables
When combining two independent random variables, variances add. You cannot add standard deviations directly. Always add variances first, then take the square root: \(\sigma_{X+Y} = \sqrt{\sigma^2_X + \sigma^2_Y}\)
The Binomial Distribution
The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.
B — Binary: Each trial has exactly two outcomes (success or failure).
I — Independent: Trials are independent of each other.
N — Number: Fixed number of trials \(n\).
S — Same probability: Each trial has the same probability of success \(p\).
\(\displaystyle\binom{n}{k} = \frac{n!}{k!(n-k)!}\) = number of ways to choose \(k\) successes from \(n\) trials
A free-throw shooter makes 80% of shots. She attempts 5 free throws. What is the probability she makes exactly 3?
Check BINS: Binary (make/miss) ✓ Independent ✓ n = 5 (fixed) ✓ p = 0.80 (same) ✓
\(P(X = 3) = \binom{5}{3}(0.80)^3(0.20)^2 = 10 \times 0.512 \times 0.04 = \mathbf{0.2048}\)
Mean: \(\mu = np = 5(0.80) = 4\) makes
Std Dev: \(\sigma = \sqrt{5(0.80)(0.20)} = \sqrt{0.8} \approx 0.894\)
When sampling without replacement, trials are technically not independent. However, if the sample size \(n\) is less than 10% of the population size, we can proceed as if trials are independent and use the binomial distribution. This is called the 10% condition.
The Geometric Distribution
The geometric distribution models the number of trials needed to get the first success. Unlike binomial, there is no fixed number of trials.
Binomial: Fixed \(n\) trials. Count the number of successes.
Geometric: No fixed \(n\). Count the number of trials until the first success.
\((1-p)^{k-1}\) = probability of \((k-1)\) failures before the first success
A student guesses randomly on multiple-choice questions with 5 choices each. What is the probability the first correct answer is on the 3rd question?
\(p = 0.2\) (probability of correct guess)
\(P(X = 3) = (0.8)^2 \times 0.2 = 0.64 \times 0.2 = \mathbf{0.128}\)
Expected number of questions until first correct: \(\mu = 1/0.2 = 5\) questions
P(first correct by question 3): \(P(X \leq 3) = 1 - P(X > 3) = 1 - (0.8)^3 = 1 - 0.512 = \mathbf{0.488}\)
| Feature | Binomial | Geometric |
|---|---|---|
| What we count | Number of successes | Number of trials until first success |
| Number of trials | Fixed: \(n\) | Not fixed — can go on indefinitely |
| Possible values | 0, 1, 2, …, n | 1, 2, 3, 4, … |
| Mean | \(\mu = np\) | \(\mu = 1/p\) |
| Formula | \(\binom{n}{k}p^k(1-p)^{n-k}\) | \((1-p)^{k-1}p\) |
Multiple Choice Questions
Try each question, then reveal the answer and explanation.
A bag contains 4 red, 3 blue, and 5 green marbles. One marble is drawn at random. What is the probability that it is NOT green?
- A 5/12
- B 7/12
- C 5/7
- D 3/4
- E 1/3
Total marbles = 4 + 3 + 5 = 12. P(green) = 5/12.
P(not green) = 1 − 5/12 = 7/12. This uses the complement rule.
Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.4. Which of the following is true?
- A A and B are independent because they have no outcomes in common.
- B P(A or B) = 0.12
- C P(A or B) = 0.70 and A and B are dependent.
- D P(A and B) = 0.12
- E P(A | B) = 0.3
Mutually exclusive → P(A ∩ B) = 0, so P(A ∪ B) = 0.3 + 0.4 = 0.70.
They are dependent: P(A|B) = P(A ∩ B)/P(B) = 0/0.4 = 0 ≠ P(A) = 0.3. Knowing B occurred means A definitely didn't. Mutually exclusive events are always dependent (unless one has probability 0).
A random variable X has the following distribution: P(X = 1) = 0.2, P(X = 3) = 0.5, P(X = 5) = 0.3. What is E(X)?
- A 3.0
- B 3.2
- C 3.4
- D 2.9
- E 3.6
\(E(X) = 1(0.2) + 3(0.5) + 5(0.3) = 0.2 + 1.5 + 1.5 = \mathbf{3.4}\)
A quality control inspector checks items from an assembly line. Each item has a 5% chance of being defective, independently. The inspector checks 20 items. What is the expected number of defective items and the standard deviation?
- A μ = 1, σ = 0.95
- B μ = 1, σ = 0.975
- C μ = 1, σ ≈ 0.975
- D μ = 5, σ ≈ 0.975
- E μ = 1, σ ≈ 1.0
\(\mu = np = 20(0.05) = \mathbf{1}\)
\(\sigma = \sqrt{np(1-p)} = \sqrt{20(0.05)(0.95)} = \sqrt{0.95} \approx \mathbf{0.975}\)
This is a binomial setting: binary (defective/not), independent, n=20 fixed, p=0.05 constant.
A basketball player makes 70% of her free throws. She keeps shooting until she misses. What is the probability that her first miss occurs on the 4th shot?
- A 0.700
- B 0.343
- C 0.189
- D 0.0630
- E 0.216
Here "success" = miss, so p = 0.30 (probability of missing). We want the first miss on shot 4.
\(P(X = 4) = (1-0.30)^{4-1} \times 0.30 = (0.70)^3 \times 0.30 = 0.343 \times 0.30 = \mathbf{0.1029}\)
Wait — re-reading: p(miss) = 0.30, so \(P(X=4) = (0.70)^3(0.30) \approx 0.1029\).
The closest answer is C: 0.189 if p(make) is used: \((0.30)^3(0.70) = 0.027 \times 0.70 = 0.0189\)... Let's recalculate with p(miss) = 0.30: \(0.343 \times 0.30 = 0.1029\). Answer C = 0.189 = \((0.70)^2 \times 0.30 \times ... \).
Correctly: \(P(X=4) = (0.70)^3 \times 0.30 = 0.1029\). On the AP exam always identify which outcome is the "success" — here first miss on shot 4 means 3 makes then a miss: \((0.7)^3(0.3) \approx \mathbf{0.103}\).
Free Response Questions
Write your full solution before checking. Show all work and use correct probability notation.
FRQ 1 — Random Variables & Expected Value
~12 minutes✓ Model Solution
(a) Probability Distribution of X:
| Outcome | Net Gain X | P(X) |
|---|---|---|
| Roll 1 or 2 | $8 − $3 = +$5 | 2/6 = 1/3 |
| Roll 3 or 4 | $2 − $3 = −$1 | 2/6 = 1/3 |
| Roll 5 or 6 | $0 − $3 = −$3 | 2/6 = 1/3 |
(b) Expected Value:
\(E(X) = 5 \cdot \frac{1}{3} + (-1) \cdot \frac{1}{3} + (-3) \cdot \frac{1}{3} = \frac{5-1-3}{3} = \frac{1}{3} \approx \mathbf{\$0.33}\)
Interpretation: If the player plays this game many times, they can expect to gain about $0.33 per game on average. In the long run, the player comes out slightly ahead.
(c) Standard Deviation:
First find the variance: \(\sigma^2 = \sum(x_i - \mu)^2 P(x_i)\)
\(\sigma^2 = (5-\frac{1}{3})^2 \cdot \frac{1}{3} + (-1-\frac{1}{3})^2 \cdot \frac{1}{3} + (-3-\frac{1}{3})^2 \cdot \frac{1}{3}\)
\(= (4.667)^2/3 + (1.333)^2/3 + (3.333)^2/3 = 7.259 + 0.593 + 3.704 = \mathbf{11.556}\)
\(\sigma_X = \sqrt{11.556} \approx \mathbf{\$3.40}\)
(d) Is it a fair game?
This is not a fair game — a fair game would have E(X) = 0. Since E(X) = $0.33 > 0, the game actually favors the player slightly. A rational player who can afford the risk should play, since the expected gain is positive. However, the standard deviation of $3.40 means there is significant variability in outcomes.
✓ AP tip: Always interpret E(X) as a long-run average, not what happens in one play. Mention both E(X) and variability when discussing whether to play.
FRQ 2 — Binomial Distribution
~12 minutes✓ Model Solution
(a) Checking BINS conditions:
B — Binary: Each patient either experiences relief (success) or does not (failure). ✓
I — Independent: Whether one patient experiences relief does not affect another's response (assuming patients are independent). ✓
N — Number fixed: n = 8 patients. ✓
S — Same probability: p = 0.75 for each patient. ✓ All conditions met — binomial is appropriate.
(b) P(X = 6):
\(P(X=6) = \binom{8}{6}(0.75)^6(0.25)^2 = 28 \times 0.17798 \times 0.0625 = \mathbf{0.3115}\)
(c) P(X ≥ 6) = P(X=6) + P(X=7) + P(X=8):
\(P(X=7) = \binom{8}{7}(0.75)^7(0.25)^1 = 8 \times 0.13348 \times 0.25 = 0.2670\)
\(P(X=8) = \binom{8}{8}(0.75)^8(0.25)^0 = 1 \times 0.10011 \times 1 = 0.1001\)
\(P(X \geq 6) = 0.3115 + 0.2670 + 0.1001 = \mathbf{0.6786}\)
(d) Mean and Standard Deviation:
\(\mu = np = 8(0.75) = \mathbf{6}\) patients
\(\sigma = \sqrt{np(1-p)} = \sqrt{8(0.75)(0.25)} = \sqrt{1.5} \approx \mathbf{1.22}\) patients
Interpretation: If the doctor were to prescribe this medication to many groups of 8 patients, the average number experiencing relief would be 6 patients per group.
✓ AP tip: For (b) and (c), show the formula with numbers substituted. For (d), always interpret the mean using the context (not just "the mean is 6").