AP Statistics Quiz: Estimating Probabilities Using Simulation

What this quiz covers

This quiz focuses on Estimating Probabilities Using Simulation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The probability is exactly $0.327$ because 327 out of 1000 trials were successes.
2. About $32.7\%$ of groups of 5 students from this school are expected to have at least 2 birthdays in the same month. (correct answer)
3. In the next 1000 trials, there will be exactly 327 successes.
4. Because the random numbers 1–12 were used, the result only applies to months with 12 days.
5. The simulation shows that 327 students at the school have birthdays in the same month.

Explanation: This question assesses the AP Statistics skill of estimating probabilities using simulation, where random number generation mimics birth months to approximate the likelihood of shared birthdays in a group. The simulation repeated 1000 trials, each consisting of generating five random numbers from 1 to 12 and checking for at least two matches, resulting in 327 successes, which provides an estimate rather than an exact value. A common distractor is choice A, which incorrectly claims the probability is exactly 0.327, ignoring that simulations yield approximations that improve with more trials but are not precise. In a mini-lesson on simulation, remember that it models real-world randomness by assuming a probability model (here, equal likelihood of months) and using repeated trials to estimate outcomes, such as the proportion of successes. This approach is useful when theoretical calculations are complex, like in the birthday problem. The correct interpretation is B, as it properly states the estimate in terms of expected proportion without claiming exactness.

Question 2

1. The probability the player makes at least 8 of 10 is exactly $141/500$.
2. The player will make at least 8 of 10 in exactly 141 of the next 500 real practices.
3. The estimated probability the player makes at least 8 of 10 is about $141/500$, assuming each shot is independent with make probability $0.72$. (correct answer)
4. Because 10 random numbers were generated per trial, the probability of at least 8 makes must be $10 \times 0.72$.
5. The simulation proves the player's true make probability is $0.72$.

Explanation: In AP Statistics, estimating probabilities using simulation involves modeling events like free throws with random numbers to approximate success rates over multiple trials. Here, the coach ran 500 trials, each with 10 simulated shots where numbers 1-72 represent makes, yielding 141 trials with at least 8 makes, which estimates the probability under the assumption of independence and a 0.72 make rate. Choice A is a distractor because it treats the simulation result as the exact probability, but simulations only provide estimates that can vary. A mini-lesson on simulation: it uses random processes to replicate scenarios, repeating trials to compute proportions that approximate true probabilities, especially for binomial outcomes like this. This method assumes the underlying model is correct and helps visualize rare events. The correct choice is C, as it accurately describes the estimate with the given assumptions.

Question 3

1. The simulation shows that exactly $10\%$ of bulbs are defective.
2. The probability that a sample of 4 has at least one defective bulb is exactly $276/800$.
3. About $276/800$ is an estimate of the probability a random sample of 4 bulbs contains at least one defective bulb, assuming independence and defect rate $0.10$. (correct answer)
4. Because digits 0–9 were used, the probability depends on the number 10.
5. The next 800 samples of 4 bulbs will include exactly 276 samples with a defect.

Explanation: Estimating probabilities using simulation in AP Statistics is demonstrated here by modeling defective bulbs with random digits, where 0 represents a defect in samples of 4, repeated over 800 trials to get 276 with at least one defect. This setup assumes independence and a 0.10 defect rate, using repeated trials to approximate the complement of all good bulbs. A distractor is choice B, which asserts the probability is exact, but simulation results are estimates that could change with different runs. In a mini-lesson, simulation replicates stochastic processes by assigning outcomes to random numbers, tallying results over many trials to estimate probabilities, ideal for hypergeometric or binomial scenarios. It highlights the law of large numbers, where more trials yield better estimates. Choice C is correct, as it properly frames the result as an estimate under the assumptions.

Question 4

1. The probability is exactly $38/200$ because simulation gives the true probability when the model is correct.
2. About $38/200$ is a reasonable estimate of the probability that a day has 5 or more defectives under this model. (correct answer)
3. The probability that any single item is defective is $38/200$.
4. Because 200 days were simulated, the probability must be $38\%$ for every set of 200 real days.
5. The probability cannot be estimated without simulating every possible set of 100 items.

Explanation: This question assesses the skill of estimating probabilities using simulation in AP Statistics. The simulation involves repeated trials where each trial mimics a day of production with 100 items, each having a 0.03 chance of being defective, and counts days with 5 or more defectives across 200 trials, yielding 38 such days. This proportion, 38/200, provides an approximate estimate of the probability under the model, as simulations use many trials to approximate true probabilities through the law of large numbers. A common distractor, like choice A, incorrectly claims the probability is exactly 38/200, but simulations give estimates, not exact values, due to inherent randomness. In a mini-lesson on simulation, remember that we model real-world processes with random number generators to replicate independent events, run numerous trials to observe outcomes, and use the relative frequency of the event as our probability estimate; more trials generally improve accuracy.

Question 5

1. The probability that a bus is late is estimated to be $468/1000$.
2. The estimated probability that a day has at least 3 late buses is about $468/1000$ under the assumptions. (correct answer)
3. The probability is exactly $0.468$ because the simulation used 1000 days.
4. The simulation shows that the next day will have at least 3 late buses with probability 0.468 regardless of the assumptions.
5. The simulation is invalid because a day with 2 late buses should count as a success half the time.

Explanation: Estimating probabilities using simulation is the AP Statistics skill applied to bus lateness here. The student simulated 1000 days with 25 buses each (00-09 late for 0.10), finding 468 days with at least 3 lates, estimating about 468/1000 under the assumptions. Repeated trials simulate the process many times, using relative frequency to approximate the probability effectively. Choice A distracts by interpreting it as the single-bus probability, but the simulation estimates the group event, not individual; confusing levels is common. In a simulation mini-lesson: define probabilities (10/100 for late), use random numbers for each unit, aggregate per trial (day), check the condition, replicate extensively, and interpret the proportion as the estimate; assumptions like independence are crucial for validity.

Question 6

1. The probability is exactly 0.7844 because 10,000 trials is enough to make the result exact.
2. The probability is $\frac{10000}{7844}$ because the simulation had 7,844 successes.
3. The estimated probability of at least one 9th grader is about $\frac{7844}{10000}$ for the simulation model, and using more trials would typically yield an estimate closer to the true probability. (correct answer)
4. The simulation is invalid unless exactly 40% of the generated digits are 0, 1, 2, or 3.
5. The result guarantees that in the next 10,000 real selections of 3 students, exactly 7,844 will include at least one 9th grader.

Explanation: This question tests simulation interpretation for school selection probability. The teacher simulated 10,000 selections of 3 students where digits 0-3 represent 9th graders (40% probability), finding 7,844 trials with at least one 9th grader. This estimates the probability as 7844/10000 = 0.7844. Choice C correctly recognizes this as an estimate that would approach the true probability with more trials. Choice A wrongly claims exactness, B inverts the fraction, D misunderstands that randomness doesn't require exact 40% in finite samples, and E treats simulation as prediction. Simulation provides probability estimates through repeated random trials; the theoretical probability here is 1-(0.6)^3 = 0.784, very close to the simulation result.

Question 7

1. The probability is exactly $\frac{1022}{2000}$ because the die is fair and the simulation used a correct generator.
2. The estimated probability of at least one 6 is about $\frac{1022}{2000}$, and the estimate would typically become more accurate with many more simulated sets of 4 rolls. (correct answer)
3. The next 2,000 real sets of 4 rolls will contain exactly 1,022 sets with at least one 6.
4. The simulation is invalid unless each face 1–6 appears exactly the same number of times across all simulated rolls.
5. The probability is $\frac{2000}{1022}$ because 1,022 is the number of successes.

Explanation: This question tests basic simulation interpretation for dice probability. The student simulated 2,000 sets of 4 die rolls and found 1,022 sets with at least one 6, estimating the probability as 1022/2000 ≈ 0.511. Choice B correctly interprets this as an estimate that would become more accurate with more trials. Choice A wrongly claims exactness, C treats simulation as prediction of exact future outcomes, D misunderstands that randomness doesn't require perfect uniformity in finite samples, and E inverts the fraction. Simulation estimates probability through repeated random trials; as the number of trials increases, the estimate typically converges to the true theoretical probability (which for this problem is 1-(5/6)^4 ≈ 0.518).

Question 8

1. The probability a car runs the light is approximately $392/1500$.
2. The probability that at least 2 of 8 cars run the light is approximately $392/1500$. (correct answer)
3. Exactly 392 of the next 1500 groups of 8 cars will have at least 2 runs.
4. The simulation result must be wrong because probabilities like 0.18 cannot be simulated with 00–99.
5. Because the simulation used independence, the probability must be the same for at least 1 run and at least 2 runs.

Explanation: This question tests interpretation of simulation results for traffic modeling. The engineer correctly simulated an 18% probability by using numbers 00-17 out of 00-99 (18 numbers = 18%). She ran 1500 trials of 8 cars each, finding 392 trials where at least 2 cars ran the light. Choice A misinterprets this as a single-car probability, while C makes an exact prediction about future groups. Choice D incorrectly claims 18% can't be simulated with two-digit numbers, and E makes a nonsensical claim about independence. The simulation provides an estimate of 392/1500 for the probability that at least 2 out of 8 cars run the light, which is the compound event being studied, not the individual 18% probability.

Question 9

1. The probability of winning is exactly $0.216$ because 216 wins occurred.
2. The estimated probability of winning is about $216/1000$, and it would likely get closer to the true probability with more trials. (correct answer)
3. The die is not fair because the simulation did not produce exactly the theoretical probability.
4. The simulation shows that the sum of 3 rolls is at least 14 on exactly 216 of the next 1000 real games.
5. The simulation result means that the average sum of 3 rolls is 216.

Explanation: This AP Statistics skill of estimating probabilities using simulation applies to dice games, where summing three rolls and checking for at least 14 was simulated 1000 times, yielding 216 wins as an estimate. Repeated trials help average out variability, providing a closer approximation to the true probability of winning with a fair die. Distractor A claims exactness from the 216 wins, but simulations are empirical estimates, not theoretical certainties. Mini-lesson on simulation: it models discrete uniform distributions like die rolls by generating random values, computing outcomes per trial, and using the success proportion over repetitions to estimate probabilities. This is useful for multinomial problems where enumeration is impractical. The correct choice is B, noting the estimate and that more trials would refine it.

Question 10

1. The probability of at least 4 admissions is $4/8$ because 4 admissions are required.
2. The probability of at least 4 admissions is exactly $296/1500$ because 1500 trials is a large number.
3. About $296/1500$ is an estimate of the probability that at least 4 of 8 patients are admitted, assuming each patient's admission is independent with probability $0.30$. (correct answer)
4. The simulation means that 296 of the 1500 simulated patients were admitted.
5. The next hour with 8 patients will have at least 4 admissions with probability $0.30$.

Explanation: In AP Statistics, estimating probabilities using simulation is used here for hospital admissions, with numbers 1-3 for admission out of 10, simulated over 1500 trials of 8 patients, yielding 296 with at least 4 admissions. Repeated trials assume independence to approximate overload risks. Choice B is a distractor, asserting exactness from the large trial count, but simulations always estimate, with precision improving but not guaranteeing exactness. A mini-lesson: assign random numbers to probabilities, simulate groups, count qualifying trials, and estimate via proportion, assuming the model like 0.30 independence. This aids in resource planning. The correct interpretation is C, as it describes the estimate with the independence and probability assumptions.

Question 11

1. The probability is exactly $88/400$ because the simulation used physical cards.
2. The simulation estimates the probability of at least 2 red marbles in 4 draws without replacement to be about $88/400$. (correct answer)
3. The jar must contain exactly 88 red marbles out of 400 marbles.
4. In the long run, exactly 88 of every 400 sets of 4 draws will have at least 2 red marbles.
5. Because marbles are drawn without replacement, simulation cannot be used to estimate the probability.

Explanation: This question in AP Statistics involves estimating probabilities using simulation for hypergeometric draws without replacement, modeled by shuffling cards and drawing 4, repeated 400 times with replacement between trials, yielding 88 successes. The repeated trials, despite the without-replacement within each, provide an estimate by averaging over many shuffles. Distractor E wrongly states simulation can't be used without replacement, but it can accurately model it via physical or digital replication. Mini-lesson on simulation: for dependent events like draws without replacement, use tools like cards to mimic the population, record outcomes per trial, and estimate via proportion over repetitions. This contrasts with independent models. Choice B is correct, as it frames the result as an approximate probability estimate.

Question 12

1. The probability that a single student prefers online appointments is $271/1000$.
2. The probability that at least 7 of 12 prefer online is approximately $271/1000$. (correct answer)
3. The probability is exactly $0.271$ because 1000 trials guarantees the true probability.
4. Because 271 trials were successes, exactly 271 of the next 1000 real samples will have at least 7 online.
5. The simulation is incorrect because it should use 0–4 for "online" to represent 40%.

Explanation: This AP Statistics question tests estimating probabilities using simulation for binomial-like scenarios. The simulation runs 1000 trials, each generating 12 random digits to represent student preferences (0-3 for online, matching 40%), and finds 271 trials with at least 7 online preferences, estimating the probability as approximately 271/1000. Repeated trials help approximate the probability by observing the frequency of the event under the assumed model. A distractor like choice C claims the probability is exactly 0.271 because of 1000 trials, but simulations yield estimates that can vary; exactness requires infinite trials or direct calculation. Mini-lesson on simulation: identify the probability (0.40 here), choose a random mechanism (digits 0-9), define success (at least 7), perform many trials, and compute the relative frequency as the estimate; this method is useful when direct binomial probability calculations are tedious.

Question 13

1. The probability of at least 4 returns in 5 years is approximately $5294/10000$ under the model. (correct answer)
2. The probability of a return in any given year is approximately $5294/10000$.
3. The probability is exactly $0.5294$ because 10,000 trials eliminates all randomness.
4. Because 5294 trials were successes, the bird will return in at least 4 of the next 5 years.
5. The simulation is invalid because 00–69 includes 69 numbers, not 70.

Explanation: In AP Statistics, this question evaluates estimating probabilities using simulation for bird return patterns. The biologist simulated 10,000 sequences of 5 years (00-69 return for 0.70), finding 5294 with at least 4 returns, estimating the probability as approximately 5294/10000 under the model. Repeated trials build a distribution of outcomes, allowing the proportion to estimate the true probability effectively. Choice E is a distractor claiming invalidity because 00-69 is 70 numbers (which is correct for 70/100=0.70), but it's actually valid; miscounting codes is a common error. For a simulation mini-lesson: select a random tool (two-digit numbers), map ranges to probabilities accurately, simulate each period independently, evaluate the criterion per sequence, repeat many times, and use the success frequency as the estimate; large trials like 10,000 enhance precision.

Question 14

1. The estimated probability of winning is about $820/5000$. (correct answer)
2. The probability of winning is exactly $820/5000$ because 5000 trials is large.
3. The probability of drawing a heart on the first card is about $820/5000$.
4. Because there were 820 wins, the next 820 trials would also be wins.
5. The simulation result is invalid because shuffling changes the probability from trial to trial.

Explanation: In AP Statistics, this problem focuses on estimating probabilities using simulation for a card game scenario. The student conducted 5000 repeated trials, each simulating drawing 3 cards without replacement and checking for at least 2 hearts, resulting in 820 wins, so the estimate is about 820/5000. This approach leverages repeated simulations to approximate the winning probability, as exact calculation might be complex. Choice B is a distractor because it asserts the probability is exactly 820/5000 due to the large number of trials, but even large simulations provide approximations, not exact values, since randomness persists. For a mini-lesson on simulation: define the components (like shuffling and drawing), assign probabilities (based on deck composition), run many independent trials, record the proportion of successes, and interpret it as an estimate; here, virtual shuffling ensures each trial is fair and independent.

Question 15

1. The probability a customer tips at least 20% is estimated as $612/2000$.
2. The probability that at least 6 of 8 customers tip at least 20% is about $612/2000$. (correct answer)
3. Because 2000 is large, the probability must be exactly $0.306$.
4. The simulation result would be the same every time if repeated with 2000 trials.
5. The simulation estimates the probability that exactly 6 customers tip at least 20%.

Explanation: This AP Statistics problem involves estimating probabilities using simulation for customer tipping behavior. The student ran 2000 trials, each with 8 customers (digits 0-4 no, 5-9 yes for 0.55 probability), finding 612 groups with at least 6 yeses, so the estimate is about 612/2000. By using repeated trials, the simulation approximates the probability of the compound event under the independence assumption. A common distractor, choice C, suggests exactness due to the large sample, but simulations always provide approximations because of sampling variation. Mini-lesson on simulation: assign digits to outcomes proportionally (5/10 for yes), simulate each group's outcomes independently, check the condition per trial, conduct many trials, and calculate the proportion; this technique estimates probabilities when binomial formulas are cumbersome for 'at least' scenarios.

Question 16

1. The probability a single call is resolved is approximately $2768/10000$.
2. The probability that exactly 3 of 6 calls are resolved is approximately $2768/10000$. (correct answer)
3. Exactly 2768 of the next 10,000 sets of 6 calls will have exactly 3 resolved.
4. The probability is exactly $2768/10000$ because 10,000 trials is large enough to remove sampling variability.
5. The simulation is flawed because 0–3 represents 4 digits, which cannot model a probability of 0.4.

Explanation: This question assesses understanding of simulation interpretation for exact count probabilities. The supervisor correctly modeled a 40% resolution rate using digits 0-3 out of 0-9 (4 digits = 40%). She simulated 10,000 sets of 6 calls, counting trials where exactly 3 were resolved, finding 2768 such trials. Choice A misinterprets this as a single-call probability, while C makes a deterministic prediction. Choice D incorrectly claims large simulations eliminate variability entirely, and E wrongly suggests the digit mapping is flawed. The simulation estimates the probability of exactly 3 out of 6 calls being resolved as approximately 2768/10,000, which is a specific compound event different from the individual 40% resolution rate.

Question 17

1. The probability of at least 4 makes is exactly $\frac{4639}{8000}$ because the random-number method guarantees the true probability.
2. The estimated probability of at least 4 makes is about $\frac{4639}{8000}$ for this model, and repeating with more than 8,000 trials would likely give a more stable estimate. (correct answer)
3. The player will make at least 4 of the next 5 free throws with certainty because the simulation shows it happens more than half the time.
4. The simulation is invalid unless exactly 70% of the generated numbers are between 00 and 69.
5. The probability is $\frac{8000}{4639}$ because the simulation counted 4,639 successes.

Explanation: This question involves simulating binomial probability for free throw success. The coach simulated 8,000 sets of 5 shots using random numbers 00-99, where 00-69 represents a made shot (70% probability), finding 4,639 sets with at least 4 makes. This gives an estimated probability of 4639/8000 ≈ 0.580. Choice B correctly recognizes this as an estimate that would stabilize with more trials. Choice A wrongly claims exactness, C misinterprets probability as certainty, D misunderstands that we expect approximately 70% makes overall but not exactly, and E inverts the fraction. Simulation estimates probabilities through repeated trials of random experiments, with accuracy improving as trial count increases according to the law of large numbers.

Question 18

1. The probability of at least 3 correct is exactly $231/2000$ because guessing is random.
2. The simulation estimates the probability of at least 3 correct when randomly guessing to be about $231/2000$. (correct answer)
3. The student will get at least 3 correct on exactly 231 of the next 2000 quizzes.
4. Because there are 4 choices, the probability of at least 3 correct must be $3/4$.
5. The simulation shows the student knows 231 of the 2000 questions.

Explanation: This AP Statistics topic of estimating probabilities using simulation models quiz guessing with random 1-4 where 1 is correct, repeated 2000 times for 5 questions, producing 231 trials with at least 3 correct. The large number of repeated trials gives a reliable estimate of guessing success under the 1/4 probability. Distractor A claims exactness because guessing is random, but randomness means results are estimates varying by run. Mini-lesson: simulation for multiple trials involves generating outcomes per event, checking conditions, and using success rates over many repetitions to estimate, perfect for binomial guessing models. It demonstrates expected values empirically. Choice B is correct, properly presenting the result as an approximate probability estimate for random guessing.

Question 19

1. About $418/2000$ is a reasonable estimate of the probability of at least 3 delays in 20 packages, given a $6\%$ delay rate. (correct answer)
2. The exact probability of at least 3 delays is $418/2000$ because the simulation used 2000 trials.
3. The company will have at least 3 delays on exactly 418 of the next 2000 days.
4. The probability of at least 3 delays is $3 \times 0.06$ because 3 delays are required.
5. The simulation result means 418 packages out of 2000 are delayed.

Explanation: This AP Statistics question focuses on estimating probabilities using simulation to model package delays with a 6% rate, using random numbers to simulate daily outcomes. The manager conducted 2000 trials, each representing 20 packages with delays for numbers 00-05, resulting in 418 trials with at least 3 delays, offering an estimate of the probability. Distractor B claims the probability is exact due to the number of trials, but simulations always provide approximations, not exact values, even with large samples. Mini-lesson on simulation: it involves defining a random mechanism (like digit assignment) to mimic events, repeating trials to calculate the proportion of successes, which estimates the true probability under the model. This is particularly helpful for Poisson-like processes or when direct calculation is tedious. The correct interpretation is A, as it reasonably states the estimate with the given delay rate.

Question 20

1. The probability of at least 9 clicks is exactly $52/600$ because 600 trials were run.
2. About $52/600$ is an estimate of the probability of at least 9 clicks among 15 visitors, assuming each visitor clicks independently with probability $0.40$. (correct answer)
3. The ad will get at least 9 clicks on exactly 52 of the next 600 days.
4. The probability of at least 9 clicks is $9/15$ because 9 clicks are needed.
5. The simulation proves the click rate is greater than $40\%$.

Explanation: In AP Statistics, estimating probabilities using simulation is shown by modeling ad clicks with random numbers 1-4 for success out of 10, repeated over 600 trials of 15 visitors, resulting in 52 trials with at least 9 clicks. These repeated trials under the 0.40 click assumption allow for an empirical probability estimate. Choice A is a distractor, claiming exactness due to trial count, but simulations provide approximations subject to random variation. A mini-lesson: simulation assigns probabilities to random outcomes, replicates scenarios many times, and computes the fraction of desired events to estimate likelihoods, assuming independence as here. It's effective for binomial thresholds. The correct interpretation is B, as it describes the estimate with the independence and rate assumptions.