Lesson 14: Probability & Simulation

(a) Warm-Up

Imagine a basketball player who makes 70% of her free throws. She steps to the line for a one-and-one and then takes five shots in a stretch. What's the chance she makes at least four of those five?

You could grind this out with the binomial formula from Lesson 12 — and for five shots, you can. But suppose the question got messier: What's the probability she needs at least 8 boxes of cereal to collect all 4 different prizes hidden inside? Or In a randomly seated class of 30, what's the chance two people share a birthday? These get ugly fast. The formulas either don't exist in a tidy form or require math beyond this course.

Here's the beautiful workaround: instead of computing the probability, we imitate the random process over and over and just count how often the event happens. Flip a coin, roll a die, read random digits — anything that reproduces the right chances. Do it hundreds of times and the fraction of successes settles down near the true probability. That's simulation, and by the end of this lesson you'll be able to design one precisely enough to earn full credit on the AP exam.

(b) Core Concept

What a simulation is

A simulation is the imitation of a chance process using a model — random digits, a calculator, dice, cards — designed so that its outcomes occur with the same probabilities as the real situation. We run the model many times, count how often the event of interest happens, and use that fraction as an estimate of the true probability.

The key word is estimate. A simulation does not give you the exact probability. It gives you an approximation that gets better the more trials you run. That last idea has a name we'll return to: the Law of Large Numbers.

Why bother, if the binomial formula exists? Because most real probability questions are not binomial. As soon as outcomes pile up across stages, depend on each other, or involve a "keep going until..." rule, exact computation gets hard or impossible. Simulation sidesteps all of it. If you can describe the chance process, you can simulate it.

The five steps of every simulation design

On the AP exam, "describe a simulation" is one of the most common — and most missed — tasks. The graders want a procedure precise enough that another person could run it without asking you a single question. Every correct design has these five pieces:

1. Assign digits (or random numbers) to outcomes so the assigned probabilities match the real ones.
2. Define one trial — exactly what one repetition of the real scenario looks like.
3. State the stopping rule — when one trial ends and what counts as a "success."
4. Run many trials — the more the better.
5. Estimate the probability = (number of successes) ÷ (number of trials).

Let's build one completely, for the free-throw question from the warm-up.

A fully specified design: the 70% free-throw shooter

Question: A player makes 70% of her free throws. Estimate the probability she makes at least 4 of her next 5 shots.

Step 1 — Assign digits. We need each shot to be a "make" with probability 0.7 and a "miss" with probability 0.3. Use single random digits 0–9. Since 7 of the 10 digits should mean make, assign:

Digits 0, 1, 2, 3, 4, 5, 6  →  MAKE   (7 digits = probability 7/10 = 0.7) ✓
Digits 7, 8, 9              →  MISS   (3 digits = probability 3/10 = 0.3) ✓

Always check the count: 7 digits for make, 3 for miss, total 10. The split matches the real 0.7 / 0.3. This matching step is where most points are won or lost.

Step 2 — Define one trial. One trial = 5 shots = read 5 consecutive random digits, classifying each as make or miss.

Step 3 — Stopping rule and success. Each trial stops after exactly 5 digits. A trial is a success if it contains 4 or 5 makes (digits 0–6).

Step 4 — Run many trials. Repeat for many trials — say 30 to start, then more. Each time, read the next 5 digits and record the number of makes.

Step 5 — Estimate. Estimated P(at least 4 makes) = (number of successful trials) / (total number of trials).

Reading the results from a dotplot

Suppose we run 30 trials and, for each, plot the number of makes (0 to 5). Here is what such a run might look like:

[GRAPH: Dotplot titled "Number of Makes in 5 Shots — 30 Simulated Trials." Horizontal axis labeled "Makes" with tick marks at 0, 1, 2, 3, 4, 5. Dots stacked above each value: 0 makes → 0 dots; 1 make → 1 dot; 2 makes → 3 dots; 3 makes → 10 dots; 4 makes → 11 dots; 5 makes → 5 dots. Total = 30 dots. Center of the distribution is around 3.5 makes; clearly right-shifted because each shot is more likely to be a make than a miss.]

To estimate P(at least 4 makes), count the dots at 4 and 5: that's 11 + 5 = 16 successful trials out of 30.

Estimated P(at least 4 makes) = 16 / 30 ≈ 0.533

How good is this estimate? The exact binomial answer is

P(X ≥ 4) = C(5,4)(0.7)⁴(0.3) + (0.7)⁵ = 0.36015 + 0.16807 = 0.52822 ≈ 0.528

Our simulation gave 0.533 — strikingly close after only 30 trials. That closeness is not luck in the long run; it's the Law of Large Numbers at work.

The Law of Large Numbers

The Law of Large Numbers (LLN) says: as the number of trials increases, the proportion of successes from a simulation converges to the true probability. Ten trials might give 0.4 or 0.7 just by chance; ten thousand trials will sit tightly around 0.528.

Trials:        10      50      200     1,000   10,000
Estimate:     0.40    0.56    0.515    0.531    0.5279
True value:                   → 0.528 ←

The estimates bounce around early, then settle. Two cautions the AP exam loves to test:

• More trials → better estimate. "Too few trials" is a legitimate weakness to point out about any simulation.
• The LLN is about the long-run proportion, not about outcomes "evening out" in the short run. After three makes in a row, the next shot is still a make with probability 0.7 — the shooter has no memory. (That false belief is the "gambler's fallacy," and it is not the LLN.)

Using a random digit table

A random digit table is a long string of the digits 0–9, each equally likely and independent, usually printed in groups of five for readability:

Line 119:  71488  09984  29077  14863  61683  47052  62224  51025

The grouping is cosmetic — ignore the spaces and read straight across. To simulate, just pull digits in order and apply your assignment rule. Reading the line above in groups of 5 for the free-throw design (0–6 = make, 7–9 = miss):

71488 → makes at 1,4,8? digits 7,1,4,8,8 → make? 1 yes,4 yes → 2 makes  (not a success)
09984 → 0,9,9,8,4 → makes 0 and 4 → 2 makes                            (not a success)
14863 → 1,4,8,6,3 → makes 1,4,6,3 → 4 makes                            (SUCCESS)

Don't reuse digits. Once a digit is used in a trial, move on. Reading the same digits again for a new trial makes your trials dependent and corrupts the estimate.

Using the TI-84 randInt(

The table works beautifully when probabilities are in tenths. But what if an outcome has probability like 0.25? That's "2.5 digits" — no clean assignment exists. The calculator handles any whole-number split. The command is:

randInt(low, high, k)

It returns k random integers from low to high, inclusive, each equally likely. Find it under MATH → PROB → 5:randInt(.

• To simulate guessing on a 4-option multiple-choice question (each option equally likely), use randInt(1, 4). Let 1 = correct, 2,3,4 = wrong → P(correct) = 1/4 = 0.25, exactly. ✓
• To simulate one free throw at p = 0.7, use randInt(1, 10) and let 1–7 = make, 8–10 = miss → 7/10 = 0.7. ✓
• To simulate five shots at once: randInt(1, 10, 5) returns five integers; count how many are 7 or below.

Tip: seed your calculator first (STO→ rand, e.g. 2024 → rand) only if you need reproducible results; otherwise just run randInt(.

The logic is identical to the table — assign outcomes so the probabilities match, define a trial, set a stopping rule, run many trials, divide. The calculator just makes the random numbers and the bookkeeping faster.

(c) Worked Examples

Example 1 — Design a simulation (easy → medium)

Problem. A student guesses randomly on a 5-question quiz. Each question has 4 answer choices, exactly one correct. Describe a simulation using the TI-84 to estimate the probability the student gets at least 3 of the 5 questions correct. (Do not carry it out.)

Strategy. Each question is correct with probability 1/4 = 0.25 — not a tenth, so a single-digit table won't split cleanly. Use randInt(1, 4).

Solution.

• Assign: Use randInt(1, 4). Let 1 = correct answer (probability 1/4 = 0.25 ✓) and 2, 3, 4 = wrong (probability 3/4 = 0.75 ✓).
• One trial: Generate 5 random integers with randInt(1, 4, 5) — one per question. Count how many are 1 (correct).
• Stopping rule / success: A trial is a success if it has 3, 4, or 5 ones (at least 3 correct).
• Run many trials (e.g., 100).
• Estimate: P(at least 3 correct) = (successful trials) / (total trials).

Interpretation. For reference, the exact binomial answer is only about 0.104 — guessing your way to a good score is unlikely, and the simulation would confirm that roughly 1 in 10 simulated quizzes succeeds.

Example 2 — Carry out a simulation from a random digit table (medium)

Problem. A cereal company puts a coupon in 30% of its boxes, at random. A shopper buys 4 boxes. Using the random digit line below, run 10 trials to estimate the probability of getting at least one coupon in 4 boxes.

Line 42:  96565  75634  81625  33001  39379  47233  51298  90222

Strategy. P(coupon) = 0.30 → use 3 of 10 digits for "coupon." One box = one digit; one trial = 4 boxes = 4 digits.

Solution.

• Assign: Digits 0, 1, 2 = coupon (3/10 = 0.3 ✓); digits 3–9 = no coupon (7/10 = 0.7 ✓).
• One trial: read 4 digits; success = at least one of them is 0, 1, or 2.

Reading the line in groups of 4 (spaces ignored):

Successes = 7 out of 10 trials.

Estimated P(at least one coupon) = 7 / 10 = 0.70

Interpretation. Based on these 10 trials, we estimate about a 0.70 probability of getting at least one coupon in 4 boxes. (The exact value is 1 − 0.7⁴ = 0.7599; with only 10 trials our estimate of 0.70 is reasonable but rough — more trials would tighten it.)

Example 3 — Interpret a simulation dotplot (AP-style)

Problem. A theme park claims 40% of visitors ride the new roller coaster. A researcher simulates groups of 10 visitors, repeating the simulation 50 times, and records the number who ride. The dotplot shows the results.

[GRAPH: Dotplot titled "Number Who Ride Out of 10 Visitors — 50 Simulated Trials." Horizontal axis "Number Who Ride" with ticks 0–10. Dots above each value: 1 → 1; 2 → 4; 3 → 9; 4 → 13; 5 → 11; 6 → 7; 7 → 4; 8 → 1; 0,9,10 → 0. Total = 50 dots, centered near 4, roughly symmetric, slightly right of center mirrors the 0.4 chance.]

(i) Use the dotplot to estimate the probability that 6 or more of 10 visitors ride the coaster.

Solution. Count the dots at 6, 7, 8, 9, 10: 7 + 4 + 1 + 0 + 0 = 12 out of 50.

Estimated P(6 or more ride) = 12 / 50 = 0.24

(ii) Explain how the estimate would change if the researcher ran 5,000 trials instead of 50.

Solution. By the Law of Large Numbers, with far more trials the estimated probability would converge toward the true probability and become more stable and reliable — running again would give nearly the same value each time, rather than bouncing as a 50-trial estimate can.

(d) Common Mistakes

1. Digit assignment that doesn't match the probability. For a 35% event, a student writes "digits 1, 2, 3 = success." That's 3/10 = 0.30, not 0.35. Fix: match the proportion exactly. For 0.35, use two-digit numbers: 00–34 = success (35 of 100 = 0.35), 35–99 = failure. Always count your digits and confirm the fraction equals the real probability before going further.

2. Miscounting what one trial is. The scenario is "4 boxes," but the student reads only 1 digit per trial, or rereads 5 digits because the line is grouped in fives. Fix: one trial must reproduce the whole scenario. Four boxes = exactly four digits, regardless of how the table is spaced. Write down "one trial = ___ digits" before you start counting.

3. Too few trials. Concluding a probability from 5 or 10 trials and treating it as the truth. Fix: a small simulation gives a rough estimate. State that more trials would improve it — and if a problem asks you to critique a simulation, "only ran a few trials" is a valid weakness.

4. Reusing or skipping digits incorrectly. Starting each new trial back at the beginning of the line, or randomly jumping around. Fix: read digits once, in order, straight across the line. Each digit is used for one position in one trial, then you move on.

5. Confusing the Law of Large Numbers with "things evening out." Thinking that after several makes, a miss is "due." Fix: the LLN is about the long-run proportion converging — individual independent outcomes never become "due." Each trial is fresh.

(e) Practice Problems

A simulation is best described as:

- (A) An exact calculation of a probability using a formula

- (B) Imitating a chance process many times to estimate a probability

- (C) A method that always gives the true probability

- (D) A way to remove randomness from an experiment

An event has probability 0.40. Which single-digit assignment correctly models it?

- (A) Digits 1, 2, 3, 4 = success; 5, 6, 7, 8, 9 = failure

- (B) Digits 0, 1, 2, 3 = success; 4, 5, 6, 7, 8, 9 = failure

- (C) Digits 1, 2, 3 = success; 4–9, 0 = failure

- (D) Digits 0–4 = success; 5–9 = failure

To simulate an event with probability 0.62, the most appropriate tool is:

- (A) A single random digit, 0–9

- (B) A coin flip

- (C) Two-digit numbers 00–99 (or randInt(1,100))

- (D) A six-sided die

According to the Law of Large Numbers, increasing the number of trials in a simulation will:

- (A) Make every trial produce the same outcome

- (B) Guarantee the next outcome is a success

- (C) Move the estimated probability closer to the true probability

- (D) Have no effect on the estimate

A student simulates a process where P(success) = 0.5 and gets 5 successes in the first 5 trials. What is the probability the 6th trial is a success?

- (A) Less than 0.5, because successes are "overdue" to stop

- (B) Exactly 0.5

- (C) Greater than 0.5, because the player is "hot"

- (D) Cannot be determined

Which TI-84 command simulates rolling a fair six-sided die 4 times?

- (A) randInt(0, 6, 4)

- (B) randInt(1, 6, 4)

- (C) randInt(1, 4, 6)

- (D) randInt(4, 6)

(In context) A baseball player gets a hit 30% of the time. To estimate the probability he gets at least 2 hits in his next 4 at-bats, you assign digits 0, 1, 2 = hit and 3–9 = no hit. How many digits make up one trial?

- (A) 1

- (B) 2

- (C) 4

- (D) 10

(Reads a random-digit line) Using the assignment in Problem 7 (0,1,2 = hit) and reading the line below in groups of 4, how many hits occur in the first trial?

40027  61835  09210

• (A) 1
• (B) 2
• (C) 3
• (D) 4

A simulation of 100 trials estimates a probability as 0.61. The exact probability is later computed to be 0.58. The best explanation is:

- (A) The simulation was done incorrectly

- (B) Simulations always overestimate

- (C) Normal sampling variability; the estimate is close and would improve with more trials

- (D) The exact computation must be wrong

(In context) A spinner lands on red with probability 0.25. Which is the best way to simulate one spin?

- (A) One random digit; 0, 1 = red

- (B) randInt(1, 4); 1 = red

- (C) Two random digits; 00–25 = red

- (D) A coin flip; heads = red

11. (Short answer) An airline finds that 20% of booked passengers miss their flight. Describe how to use a random digit table to simulate whether each of 5 booked passengers shows up. Give the digit assignment and define one trial.

12. (Short answer) Explain, in one or two sentences, the difference between the Law of Large Numbers and the (false) "gambler's fallacy" idea that outcomes even out in the short run.

13. (Short answer — reads a random-digit line) A free-throw shooter makes 60% of her shots. Using 0–5 = make, 6–9 = miss, and reading the line below in groups of 5, carry out 4 trials and estimate the probability she makes all 5 shots in a set of 5.

03474  37386  36964  73661  46986

14. (Short answer) A student claims, "I ran my simulation 8 times and got the answer 0.375, so the probability is 0.375." Identify the two weaknesses in this claim.

15. (Short answer — design) A teacher wants to estimate the probability that, in a class of 25 students, at least two share a birthday (ignore leap years; assume 365 equally likely birthdays). Describe how you would use the TI-84 randInt( to simulate one trial, and state the success condition.

(f) FRQ Practice (10 points)

Statistical Practice 3 — Analyze Data

A tennis player's first serve goes in 60% of the time, independently from serve to serve. In a particular game she will hit 5 first serves. Her coach wants to estimate the probability that at least 4 of those 5 first serves go in.

(a) Describe a simulation, using a random digit table, to estimate this probability. Your description must include the digit assignment, the definition of one trial, and what counts as a successful trial. (4 points)

(b) Carry out 10 trials of your simulation using the random digit line below, reading the digits in order. Show your work by recording, for each trial, the number of serves that go in and whether the trial is a success. (4 points)

Line 137:  03474  37386  36964  73661  46986  37162  33261  68045  60111  41095

(c) Use your 10 trials to estimate the probability that at least 4 of her 5 first serves go in, and interpret this estimate in context. (2 points)

Model Response

(a) Design.

• Assign digits. Each serve goes in with probability 0.60, so 6 of the 10 digits mean "in":

`

Digits 0, 1, 2, 3, 4, 5 → serve goes IN (6/10 = 0.60 ✓)

Digits 6, 7, 8, 9 → serve is OUT (4/10 = 0.40 ✓)

`

• One trial. Read 5 consecutive random digits (one per serve) and count how many are 0–5 ("in").
• Successful trial. A trial is a success if 4 or 5 of the digits are "in" (digits 0–5).
• Run many trials and estimate P = successes ÷ trials.

(b) Carry out 10 trials (each group of 5 digits = one trial; "in" = 0–5):

Successes: trials 1, 7, 9, 10 → 4 successes out of 10.

(c) Estimate and interpret.

Estimated P(at least 4 of 5 first serves in) = 4 / 10 = 0.40

Based on these 10 simulated trials, we estimate the probability that at least 4 of the player's 5 first serves go in is about 0.40. (Note: 10 trials is a small simulation; the estimate would become more reliable with many more trials. For reference, the exact binomial value is 0.337.)

Rubric (10 points)

Part (a) — Design (4 points)

• 1 pt: Correct digit assignment for "in" — 6 of 10 digits (e.g., 0–5) → 0.60. (Any valid 6-digit set earns the point.)
• 1 pt: Correct digit assignment for "out" — the remaining 4 digits → 0.40, consistent with the "in" set.
• 1 pt: One trial correctly defined as reading 5 digits (one per serve).
• 1 pt: Success condition correctly stated as at least 4 "in" (4 or 5 of the digits).

Part (b) — Carry out (4 points)

• 1 pt: Reads digits in order, in groups of 5, without reusing or skipping.
• 1 pt: Correctly classifies each digit as in/out per the assignment.
• 1 pt: Correctly counts "# in" for each trial.
• 1 pt: Correctly identifies which trials are successes (4 of 10 with this line/assignment).

Part (c) — Estimate and interpret (2 points)

• 1 pt: Computes the estimate consistent with part (b): successes ÷ 10 (here 4/10 = 0.40).
• 1 pt: Interprets in context as an estimated probability that at least 4 of 5 first serves go in (a correct contextual probability statement; acknowledging it is an estimate is good practice).

Total: 10 points.

Where students lose points

• Mismatched digits (a): assigning 7 digits (or 5) to "in" instead of 6 — the proportion no longer equals 0.60. Costs both assignment points.
• Wrong trial length (a/b): treating each digit as a full trial, or reading only 4 digits because they forgot it's 5 serves.
• Vague success rule (a): writing "she serves well" instead of the precise "4 or 5 of the 5 digits are 0–5."
• No context in (c): reporting "0.40" with no sentence tying it to first serves going in. The interpretation point requires context.
• Arithmetic slips in (b): miscounting "in" digits — careful tallying is the whole point of part (b).