Free math worksheets

- Count how many equally likely digits are assigned to the event. - Compare that fraction of the ten digits with the target probability. - A valid mapping must give the event exactly the intended share of equally likely outcomes.

1. Four of the ten equally likely digits are assigned to the event. 2. The simulated event probability is therefore \(\frac{4}{10}=0.40\). 3. This matches the required probability, so the mapping is valid.

Yes. Four of the ten equally likely digits represent the event, so its simulated probability is \(\frac{4}{10}=0.40\).

- A simulation estimate uses the proportion of trials in which the event occurred. - Put the number of successes over the total number of trials. - Express the resulting fraction as a decimal or percent.

1. Use the simulated relative frequency of success. 2. There were \(7\) successes in \(20\) trials, so the estimate is \(\frac{7}{20}=0.35\).

The estimated probability of success is \(0.35\), or \(35\%\).

- A simple random sample must not select the same population member twice. - For an odd sample size, identify which ordered position is the median. - Define one unambiguous success indicator before repeating the trial.

1. Randomly select \(5\) distinct population labels from the \(9\), so every five-element subset is equally likely. 2. Sort the five selected values and identify the third value, which is the sample median. 3. Record a success when that third value is greater than \(15\). 4. Repeat many trials and use the successful-trial proportion as the estimate.

Select \(5\) distinct values uniformly, sort them, and record a success when the third value exceeds \(15\). The long-run success proportion estimates the probability.

- The physical process does not replace a token within one trial. - Independent repetitions require restoring the original urn. - Define one clear success indicator for each three-token sample.

1. Represent the urn with \(8\) distinct labels, \(5\) blue and \(3\) gold. 2. For each trial, randomly select \(3\) distinct labels and record success when exactly \(2\) are blue. 3. Return all labels before the next trial; the success proportion estimates the probability.

Randomly select \(3\) distinct tokens from a reset \(5\)-blue, \(3\)-gold urn for each trial. Record success for exactly \(2\) blue tokens, and use the long-run success proportion.

- Represent each person by one month label. - The event depends on repeated categories, not a particular month. - One repetition must include all \(8\) simulated people.

1. Generate \(8\) independent integers from \(1\) through \(12\), one for each person. 2. A trial is successful if any month label appears at least twice. 3. Repeat many trials; the proportion of successful trials estimates the probability.

Generate \(8\) month labels per trial and record success when a repeated label occurs. The success proportion estimates the probability.

- Combine all stopping categories that satisfy the first event. - Use the total number of repetitions as the denominator. - Treat “no success” as its own recorded outcome.

1. First success by attempt \(3\) occurred \(390+310+252=952\) times. 2. Its estimated probability is \(\frac{952}{2000}=0.476\). 3. No success has estimated probability \(\frac{552}{2000}=0.276\).

The estimates are \(0.476\) for first success by attempt \(3\) and \(0.276\) for no success in \(6\) attempts.

- Which aspect of a sequence matters for a run but disappears in a simple count? - Decide exactly what summary should be recorded after each trial. - Estimate the probability from the fraction of trials whose recorded summary meets the cutoff.

1. Generate an ordered sequence of \(12\) independent fair-coin outcomes. 2. Scan the sequence for consecutive heads and record the maximum run length. 3. The estimate is \(\frac{3118}{10{,}000}=0.3118\). 4. The total number of heads does not preserve their order, so it cannot determine the longest run.

The estimated probability is \(0.3118\), or \(31.18\%\). The ordered sequence must be retained because the total number of heads alone does not determine the longest run.

- Track distinct categories rather than just the number of draws. - Repeated labels do not increase the number of types collected. - One trial ends after exactly \(7\) boxes.

1. Generate \(7\) independent labels from the four puzzle types. 2. Record success if all four distinct labels appear at least once. 3. The number of boxes is fixed at \(7\); the simulation must retain which types appeared, not just the count of draws.

Generate \(7\) puzzle-type labels per trial and record success when the set of observed labels contains all four types. Recording only the number of boxes is insufficient because every trial uses \(7\) boxes; the distinct type labels determine success.

- Think about what resetting a generator’s starting state does to the next values it produces. - Distinguish making the whole simulation reproducible from restarting each individual trial. - Consider whether the generator state should advance from one trial to the next.

1. Setting the same seed resets the pseudorandom generator to the same starting state. 2. Because the reset occurs before every trial, each trial repeats the same pseudorandom sequence instead of advancing to new values. 3. To make the overall run reproducible while allowing distinct trials, set the seed once before the full simulation and then let the generator state advance throughout the trials.

Resetting the same seed before every trial restarts the same pseudorandom sequence, so the trials repeat. Set the seed once before the full simulation run, not inside the trial loop.

- Use one consistent numerical scale for both start times. - Express overlap in terms of the distance between the starts. - Decide how an exact endpoint contact should be treated before running trials.

1. Represent each start time by minutes after \(1{:}00\) p.m. and generate two independent uniform values \(X,Y\) in \([0,120]\). 2. Two \(30\)-minute appointments overlap when their start times differ by less than \(30\) minutes. 3. Record success when \(|X-Y|<30\). 4. Repeat many trials and use the success proportion.

Generate two independent start times \(X,Y\) uniformly from \([0,120]\) minutes after \(1{:}00\) p.m. Record a success when \(|X-Y|<30\); the success proportion estimates the overlap probability.

- Convert each success count into a simulated relative frequency. - Compare the resulting output probabilities, not just the two input values of \(p\). - A sensitivity comparison is clearest when one modeling assumption changes at a time.

1. The estimated at-least-one probabilities are \(671/1000=0.671\) and \(834/1000=0.834\). 2. The estimates differ by \(0.834-0.671=0.163\), or \(16.3\) percentage points. 3. That output change is substantial relative to the decision quantity, so the simulation does not support the claim that the model choice cannot matter much. 4. The trial structure, number of opportunities, success definition, and simulation size should be held fixed; the uncertain probability \(p\) should be the modeled feature that changes.

The estimated probabilities are \(0.671\) and \(0.834\), a difference of \(0.163\), or \(16.3\) percentage points. The simulation therefore shows meaningful sensitivity to the assumed value of \(p\). Keep the trial structure and recorded event fixed while changing only the model probability being tested.

- A positive model probability should correspond to reachable generator outcomes. - Inspect both outcome assignment and result recording. - Use targeted test inputs rather than relying only on random trials.

1. Category D may have no random-number codes assigned, or its condition may be unreachable because of an incorrect boundary. 2. A label may also be overwritten or omitted when results are tallied. 3. Inspect the mapping and test representative random values from D’s intended interval.

Likely causes include an empty/unreachable code range or a tallying error. Check the mapping boundaries with test values that should produce category D.

- Identify which trial detail distinguishes the two requested events. - Final outcomes can discard timing information. - Store the smallest state that supports both later calculations.

1. A final success indicator does not reveal when success first occurred. 2. The two requested events require the stopping step or first-success step. 3. Store the first-success step for each trial, including a code for no success by step \(5\).

The binary log is insufficient. Store the first-success step, with a separate value for no success by step \(5\).

- What four counts summarize the allocation after all balls have been assigned? - Express the event in terms of the largest of those counts. - Use the proportion of trials satisfying the capacity condition.

1. Generate \(12\) independent bin labels, each uniformly selected from \(1,2,3,4\). 2. Count how many labels of each type occur and find the largest bin count. 3. Record success when that maximum is at most \(4\). 4. The estimate is \(\frac{3126}{8000}=0.39075\).

The estimated probability is \(0.39075\), or \(39.075\%\).

- Separate variation caused by randomness from a change in the simulation method. - To combine runs, combine their success counts and their trial counts. - The final estimate is one overall simulated relative frequency.

1. Random simulation results vary from run to run because different random outcomes can occur in each set of trials. 2. Across the four runs there are \(42+51+47+44=184\) successes in \(400\) trials. 3. The combined estimate is \(\frac{184}{400}=0.46\).

a) Random variation makes repeated simulation estimates differ even when the procedure is unchanged. b) \(0.46\)

- First convert each success count to a simulated relative frequency. - Do not assume that the larger displayed estimate is closer to an unknown true probability. - Compare the trial counts when deciding which design should fluctuate less across repetitions.

1. Run A estimates \(\frac{31}{50}=0.62\), while Run B estimates \(\frac{1190}{2000}=0.595\). 2. Because the true probability is unknown, the fact that one displayed estimate is larger does not show that it is closer to the truth. 3. With more independent trials, random highs and lows tend to average out more, so the simulated relative frequency has less run-to-run variability. 4. Therefore the \(2000\)-trial design would generally produce the more stable estimates.

a) Run A: \(0.62\); Run B: \(0.595\). b) The \(2000\)-trial design would generally vary less from run to run because larger numbers of independent trials produce more stable simulated relative frequencies.

- Count unordered selections with combinations. - For exactly \(3\) matches, choose matched numbers from the player's selections and unmatched numbers from the remaining numbers. - Relative frequency is the number of successful trials divided by the total number of trials.

1. There are \(\binom{49}{6}=13{,}983{,}816\) possible drawings. To obtain exactly \(3\) matches, choose \(3\) of the player's \(6\) numbers and \(3\) of the other \(43\) numbers. Thus, \(P(\text{exactly 3 matches})=\frac{\binom{6}{3}\binom{43}{3}}{\binom{49}{6}}=\frac{246820}{13983816}\approx 0.0176504\). 2. The relative frequency is \(\frac{17612}{1000000}=0.017612\). The absolute difference is \(\left|0.017612-0.0176504\right|\approx 0.0000384\), or about \(0.00384\) percentage points. 3. The law of large numbers states that, as the number of independent trials increases, the relative frequency tends to stabilize near the theoretical probability. The small difference after one million trials is consistent with this behavior.

1. \(P(\text{exactly 3 matches})\approx 0.0176504\), or about \(1.765\%\) 2. Relative frequency \(0.017612\); absolute difference approximately \(0.0000384\), or \(0.00384\) percentage points 3. With many trials, the relative frequency tends to approach the theoretical probability.

- Count how many input digits produce each reported result. - A fair mapping needs the same number of accepted generator outcomes per die face. - Rejected digits do not count as completed trials.

1. Digits \(0\) and \(6\) produce result \(1\); \(1\) and \(7\) produce \(2\); \(2\) and \(8\) produce \(3\); \(3\) and \(9\) produce \(4\). Results \(5\) and \(6\) each have one producing digit. Thus the probabilities are \(0.20,0.20,0.20,0.20,0.10,0.10\), not \(1/6\). 2. Accept digits \(0\) through \(5\), map them to die results \(1\) through \(6\), and reject digits \(6\) through \(9\). Generate again after a rejection. 3. Four of ten digits are rejected, so the expected rejection proportion is \(0.40\).

a) The simulated probabilities are \((0.20,0.20,0.20,0.20,0.10,0.10)\), so the method is biased. b) Accept digits \(0\) through \(5\) and map them one-to-one to results \(1\) through \(6\); reject digits \(6\) through \(9\). c) The expected rejection proportion is \(0.40\).

- Compare which random outcomes two neighboring trials share. - Independent trials should not reuse nearly all of the same random inputs. - Divide the available sequence into disjoint blocks.

1. Consecutive windows share \(9\) of their \(10\) flips, so neighboring trial indicators are strongly linked. 2. Treating the \(991\) overlapping windows as independent exaggerates the effective amount of simulation information. 3. Partition the sequence into \(100\) nonoverlapping blocks: flips \(1\) through \(10\), \(11\) through \(20\), and so on. Alternatively, generate a fresh independent block for every trial. 4. The corrected design has \(100\) trials when restricted to the given \(1000\) flips.

The overlapping windows are dependent because adjacent trials share \(9\) flips. Use \(100\) nonoverlapping blocks of \(10\) flips, or generate independent blocks separately.

- Preserve the fixed bracket structure while randomizing team locations. - Ask what must be true about the teams’ halves for a final-round meeting to be possible. - A trial need not simulate game winners if only bracket placement is being studied.

1. Generate a uniformly random permutation of the eight team labels and place them into the eight fixed bracket slots. 2. Determine which half of the bracket contains A and which half contains B. 3. Record success when they are in opposite halves; only then can their first possible meeting be the championship. 4. Repeat many independent bracket assignments and use the success proportion.

Randomly permute all eight teams into the fixed bracket and record a success when A and B occupy opposite halves. The success proportion estimates the probability that their first possible meeting is the championship.

- One random digit gives ten equally likely outcomes. - Count how many digits are assigned to option A and divide by \(10\). - Compare the distances from \(0.55\) to the attainable tenths.

1. With \(8\) event digits, Yuna's rule has probability \(\frac{8}{10}=0.8\). 2. A one-digit rule can produce probabilities only in increments of \(0.1\). Both \(0.5\) and \(0.6\) are \(0.05\) away from \(0.55\), so both are closest. 3. One valid improved rule is to assign any \(6\) distinct digits to option A, giving probability \(\frac{6}{10}=0.6\). Assigning \(5\) digits, giving \(0.5\), is equally close.

Yuna's rule gives \(0.8\), which is too high. A better one-digit rule may use \(6\) of the ten digits, giving \(0.6\); using \(5\) digits, giving \(0.5\), is equally close to \(0.55\).

- Compare interval length with the number of discrete listed values. - A single point behaves differently in continuous and discrete models. - Identify the exact output set of the generator before coding the event.

1. The continuous interval has length \(0.30\), so its probability is \(0.30\). 2. A single endpoint has probability \(0\) in a continuous uniform model, so including \(0.30\) changes nothing. 3. For outputs restricted to \(0.00,0.01,\ldots,0.99\), including \(0.30\) adds one discrete outcome and changes the probability.

The continuous probability is \(0.30\). Endpoint inclusion does not matter continuously, but it can add an outcome in a discrete hundredths generator.

- Convert two locations on a line into the lengths of the three consecutive pieces. - What ordering makes those lengths well defined? - Use the stated condition on the largest piece as the trial’s success rule.

1. Generate two independent uniform numbers \(U\) and \(V\) in \([0,1]\), then set \(a=\min(U,V)\) and \(b=\max(U,V)\). 2. The three piece lengths are \(a\), \(b-a\), and \(1-b\). 3. Record success when the maximum of these three lengths is less than \(\frac12\). 4. Sorting is required so the differences represent nonnegative consecutive segment lengths.

Generate two uniform break locations, sort them as \(a<b\), compute \(a\), \(b-a\), and \(1-b\), and record a success when all three are less than \(\frac12\). The success proportion estimates the probability.

- Identify the complete set that satisfies the condition. - Ask whether each retained outcome had the same probability before conditioning. - Compare the acceptance rate with a method that samples only valid outcomes.

1. The accepted ordered pairs are \((4, 6),(5, 5),(6, 4),(5, 6),(6, 5),(6, 6)\). 2. In rejection sampling, every original ordered pair has probability \(\frac{1}{36}\); conditioning on acceptance makes the six accepted pairs equally likely. 3. Directly selecting one of the six pairs therefore produces the same conditional distribution. 4. Direct selection always completes in one attempt, while rejection accepts only \(\frac{6}{36}=\frac{1}{6}\) of raw rolls and needs \(6\) attempts on average.

Both methods are uniform over \((4, 6),(5, 5),(6, 4),(5, 6),(6, 5),(6, 6)\). Direct selection uses \(1\) attempt, while rejection sampling uses \(6\) attempts on average.

- The next decision depends on the current position. - Continue until one of the two absorbing boundaries is reached. - Reset the position to \(0\) for every new trial.

1. Start each trial at position \(0\). 2. Generate each move with probabilities \(0.60\) and \(0.40\), update the running position, and stop at either boundary. 3. Record success for boundary \(+3\); the success proportion estimates the desired probability.

Simulate the running position step by step, stopping at \(+3\) or \(-2\), and record success when \(+3\) is reached first.

- Relate the fraction of random points in a region to the fraction of the square’s area occupied by that region. - What geometric condition identifies whether a point lies inside the quarter circle? - After estimating one quarter of the circle, account for the full circle.

1. The unit square has area \(1\), so the hit proportion estimates the quarter-circle area. 2. The estimated area is \(\frac{39{,}241}{50{,}000}=0.78482\). 3. A quarter circle of radius \(1\) has one fourth the area of the full circle, so \(\pi\) is estimated by \(4\cdot0.78482=3.13928\).

Estimated quarter-circle area: \(0.78482\). Estimated \(\pi\): \(3.13928\).

- The second-stage probability depends on the first-stage category. - Preserve both the group label and test result in the simulation record. - The final denominator is the set of simulated positives, not all trials.

1. First simulate risk group with probabilities \(0.20\) and \(0.80\). 2. Conditional on the simulated group, generate the positive/negative result using that group’s rate. 3. Among all simulated positive results, divide the high-risk positives by total positives.

Simulate risk group first, then test result using the corresponding conditional rate. Estimate the desired proportion as high-risk positives divided by all positives.

- Ask what independence would imply about sensor 2’s probability after learning sensor 1’s result. - Compare the two supplied conditional probabilities before judging the model. - A shared object-level characteristic can create dependence between two measurements of the same object.

1. If the two detections were independent, knowing whether sensor 1 detected the flaw would not change the probability that sensor 2 detects it. 2. The conditional probabilities \(0.82\) and \(0.31\) are very different, so the pilot data show substantial dependence between the sensor outcomes. 3. A better simulation could include a shared object-level feature such as flaw severity and then use severity-dependent detection probabilities for both sensors.

Independent Bernoulli trials are not supported by the pilot data because sensor 2’s detection probability changes from \(0.31\) to \(0.82\) depending on sensor 1’s outcome. A better simulation could first simulate a shared feature such as flaw severity and condition both sensor outcomes on it.

- Circular arrangements that differ only by rotation represent the same seating. - Think about choosing one fixed reference position before randomizing the other people. - Remember that adjacency wraps around the circle.

1. Fix Ava in the designated reference seat and uniformly permute the other seven people among the remaining seats. This generates each circular arrangement exactly once. 2. Ben is adjacent to Ava precisely when Ben occupies either of the two seats next to Ava, including the wraparound neighbor. 3. Every circular arrangement has exactly one rotation that places Ava in the designated seat. Fixing Ava therefore selects one representative from each rotation class without changing relative likelihoods.

a) Fix Ava at the reference seat and uniformly permute the other seven people. b) Record success when Ben is in either seat neighboring Ava. c) Fixing Ava removes rotational duplicates; it does not favor any circular arrangement.

- Ask whether the discarded trials are a random subset of all initiated trials. - Keep the target quantity in view: the question concerns what eventually happens, not what happens by minute \(30\). - A valid rule must give every initiated trial a terminal outcome before it enters the estimate.

1. Discarding unfinished trials removes a nonrandom subset of customers whose behavior is related to decision time. 2. If long-decision customers have a different eventual purchase rate, the completed trials are not representative of all initiated trials. 3. Because the target is eventual purchase, every trial should continue until the customer reaches one of the two terminal outcomes: purchase or leave without purchasing. 4. The estimate should then use all initiated trials in its denominator.

Discarding the unfinished trials can bias the estimate because the discarded customers are selected according to behavior related to the outcome process. Continue every simulated customer to purchase or departure, and estimate the eventual-purchase probability from all initiated trials.

- Use one common random input for each model pair. - The larger-threshold model succeeds in every trial where the smaller one succeeds. - Pairing focuses attention on cases where predictions differ.

1. Generate \(U\) uniformly on \([0,1)\). 2. Model 1 succeeds when \(U<0.40\); Model 2 succeeds when \(U<0.50\). 3. Shared random numbers make trial-by-trial differences attributable to the model thresholds and can reduce noise in the comparison.

Use the same \(U\) for both models, with thresholds \(0.40\) and \(0.50\). Pairing reduces irrelevant simulation variation when comparing the models.

- Treat the four joint cells as the simulation outcomes, not the two components separately. - Allocate a number of equally likely integers proportional to each joint probability. - Compare the specified intersection with the product of the marginal probabilities.

1. One valid mapping is: \(00\) through \(07\) for both fail, \(08\) through \(13\) for only A fails, \(14\) through \(17\) for only B fails, and \(18\) through \(99\) for neither fails. 2. The marginal failure probabilities are \(P(A\text{ fails})=0.08+0.06=0.14\) and \(P(B\text{ fails})=0.08+0.04=0.12\). 3. Independent marginal simulations would give \(P(\text{both fail})=0.14\cdot0.12=0.0168\), not \(0.08\). The single joint mapping preserves the dependence encoded in the table.

a) For example, map \(00\) through \(07\) to both fail, \(08\) through \(13\) to only A, \(14\) through \(17\) to only B, and \(18\) through \(99\) to neither. b) \(P(A\text{ fails})=0.14\), \(P(B\text{ fails})=0.12\). c) Independent simulation would give only \(0.0168\) for both failures instead of \(0.08\).

- The condition fixes the number of rainy days before the event is checked. - A direct trial should generate only schedules allowed by the condition. - Translate consecutive rainy days into a condition on selected day labels.

1. Select four distinct day labels uniformly from \(\{1,2,\ldots,10\}\) and mark exactly those days rainy. Equivalently, uniformly shuffle four R labels and six N labels. 2. Sort the four rainy day numbers and record success if any two consecutive selected numbers differ by \(1\). 3. Independent Bernoulli outcomes allow any rain count from \(0\) to \(10\). The stated model restricts the sample space to schedules with exactly four rainy days and gives those schedules equal probability.

a) Uniformly choose four distinct days from the ten and mark them rainy. b) Record success when two selected day numbers differ by \(1\). c) Independent \(0.4\)-probability days do not enforce exactly four rainy days, so they use a different sample space.

- Estimate the event rate separately in each stratum. - Use population proportions, not simulation sample proportions, as weights. - Distinguish allocation for precision from prevalence in the target population.

1. Estimated event rates are \(\frac{126}{1000}=0.126\) for high risk and \(\frac{31}{1000}=0.031\) for low risk. 2. Weight these rates by the population shares: \(0.20\cdot0.126+0.80\cdot0.031=0.0500\). 3. The unweighted pooled rate \(\frac{157}{2000}=0.0785\) treats the two strata as equally common. 4. Equal simulation allocation was a design choice and must not replace the population weights.

The stratified estimate is \(0.0500\). The raw pooled rate \(0.0785\) is inappropriate because the simulation used equal stratum sizes while the population proportions are \(0.20\) and \(0.80\).

- Maintain a cumulative state after every roll. - The stopping rule ends the trial at the threshold. - Positive increments make an overshoot irreversible.

1. Start each trial with total \(0\) and add independent die results. 2. Stop at the first total at least \(12\). 3. Record success only for total \(12\); a total above \(12\) cannot return to \(12\) because all later additions are positive.

Simulate a running total until it first reaches at least \(12\). Count success only when the stopping total is exactly \(12\).

- Build one trial from the actual games that make up the tournament. - Each team’s record must come from the same shared set of game outcomes. - Think about constraints that every legitimate collection of five team win totals must satisfy.

1. There are \(\binom{5}{2}=10\) games. For each unordered pair, independently generate one fair binary outcome identifying the winner. 2. Tally wins from the ten shared game outcomes. The trial succeeds if one team wins all four of its games. 3. Team win totals are constrained by shared games: every game contributes one win to one team and one loss to another, and the total number of wins must be \(10\). Independently generated team totals can violate these constraints and need not correspond to any tournament.

a) Independently simulate one fair winner for each of the \(10\) team pairs. b) Record success if a team has \(4\) wins. c) Team win totals cannot be generated independently because teams share games and the totals must be jointly consistent.

- Generate each physical link once per trial rather than simulating routes separately. - A trial needs only one overall connectivity result, even if several routes are available. - Estimate reliability from the fraction of complete network configurations that connect \(S\) to \(T\).

1. Generate one working/failed status for each of the five physical links. 2. On that single five-link configuration, determine whether \(S\) and \(T\) are connected. The diagram has four simple routes: links \(1\)-\(2\), \(3\)-\(4\), \(1\)-\(5\)-\(4\), and \(3\)-\(5\)-\(2\). 3. Record one success if at least one route works, even when several routes work. 4. The simulated reliability is \(\frac{22790}{25000}=0.9116\), or \(91.16\%\).

Generate one status for each physical link and record a single success whenever the resulting network connects \(S\) to \(T\). The estimated reliability is \(0.9116\), or \(91.16\%\).

- How should a claim of fairness be interpreted for a finite run rather than an infinite sequence? - Compare the deviations with the scale of the full sample before deciding what they establish. - What additional evidence would help distinguish ordinary random fluctuation from a persistent problem with the generator?

1. Counts near \(1000\) are expected to vary randomly in a finite sample. 2. The displayed deviations alone do not prove the generator is biased. 3. A reasonable next step is a formal goodness-of-fit check or repeated larger runs while inspecting the generator implementation.

No. Finite random variation can produce these differences. Use a formal distribution check and additional runs before concluding bias.

- Compare the probability of falling within radius \(r\) with the fraction of total area there. - Build the radius from the desired cumulative distribution. - The angle and radius must be generated independently.

1. Uniform area in a disk requires \(P(R\le r)\) to equal the fraction of disk area inside radius \(r\), which is \(\pi r^2/\pi=r^2\). Setting \(R=U\) instead gives \(P(R\le r)=r\), which is too large for \(0<r<1\), so points concentrate near the center. 2. Set \(U=R^2\), giving \(R=\sqrt{U}\). Then \(P(R\le r)=P(U\le r^2)=r^2\). 3. Generate independent \(U,V\), use \(R=\sqrt{U}\) and \(\Theta=2\pi V\), then convert to \((R\cos\Theta,R\sin\Theta)\) if Cartesian coordinates are needed.

a) Uniform radius gives too much probability to small disks because disk area grows with \(r^2\), not \(r\). b) Use \(R=\sqrt{U}\). c) Generate independent \(U,V\sim\text{Uniform}(0,1)\), set \(R=\sqrt{U}\), \(\Theta=2\pi V\), and use the resulting polar point.

- Ask which random fluctuations cause the procedure to stop and which cause it to continue. - Separate a rule about sample size or precision from a rule about obtaining a preferred result. - A defensible protocol should be specified before the simulated outcomes are seen.

1. The procedure preferentially stops after an upward random fluctuation. Runs whose estimates stay below \(0.60\) continue, while favorable fluctuations are frozen and reported, so the stopping rule selects high estimates. 2. A valid fixed-sample plan chooses the number of trials before observing outcomes and reports successes divided by that fixed total. 3. For an outcome-neutral precision rule, choose a target Monte Carlo standard error \(\varepsilon\) in advance and stop at the first batch size \(n\) satisfying \(1/(2\sqrt n)\le\varepsilon\). Since \(\sqrt{p(1-p)/n}\le1/(2\sqrt n)\) for every probability \(p\), this rule depends only on the prespecified precision target and total number of trials, not on whether the current estimate is high or low.

The reported estimate can be biased upward because stopping is triggered by an unusually high cumulative proportion. Use a prespecified trial count, or stop at the first batch size that meets a prespecified worst-case precision bound such as \(1/(2\sqrt n)\le\varepsilon\); neither rule favors high estimates.

- Model actions in the order they occur. - Identify which doors the host is allowed to open after earlier outcomes are known. - Do not replace a conditional host action with an unrestricted random choice.

1. Randomly assign the prize door and the contestant’s initial choice. 2. The host selects uniformly from the unchosen nonprize doors; when only one is available, that door is forced. 3. Record whether staying wins and whether switching to the remaining closed door wins. 4. Allowing the host to reveal the prize creates outcomes that the actual process forbids and changes the conditional structure of the game.

Simulate the prize and initial choice first, then choose the host’s door only from eligible unchosen nonprize doors. A host rule that can reveal the prize is not the same probability model.

- What information from one minute affects the next minute? - Decide a consistent order for events that can occur during the same minute. - Consider two trials with equal totals but different timing; could their maximum lines differ?

1. The trial must track at least whether the clerk is busy, the number waiting, and whether the line has ever reached \(5\). 2. In each minute, first simulate the possible arrival and update the line or clerk status; then simulate a completion if the clerk is busy and move the next waiting customer into service when appropriate. 3. Record success if the waiting-line count reaches \(5\) at any time. 4. Totals alone omit the timing and order that determine queue buildup.

Track the clerk’s status, current queue length, and whether the queue has reached \(5\). Update arrivals before end-of-minute completions, and record a success whenever the waiting count reaches \(5\). Aggregate arrival and completion totals are insufficient because their timing determines the queue.

- Read the two branches that belong to the machine’s current state. - The state reached on one cycle determines which probabilities apply on the next cycle. - One simulation trial is one six-transition path.

1. Start in the normal state. 2. At each cycle, use the transition probabilities listed under the current state in the tree and update the state. 3. After the sixth transition, record whether the machine is warm. 4. Six independent trials with one fixed warm probability are inappropriate because the next-state probabilities depend on the current state.

Simulate six state transitions sequentially, each time using the rule for the machine’s current state. Record success when the state after cycle \(6\) is warm.

- Compare the distribution immediately after the fixed start with the distribution expected after many transitions. - Ask whether the starting state still has a strong influence after only one transition. - A burn-in period means letting the same state process evolve before collecting measurements.

1. From the inactive state, the one-step probability of becoming active is \(0.20\), so a one-transition estimate is strongly tied to the chosen starting state. 2. For the long-run active proportion \(\pi_A\), balance the next-step active probability: \(\pi_A=0.20(1-\pi_A)+0.90\pi_A\). 3. Solving gives \(\pi_A=\frac{2}{3}\). 4. A burn-in approach runs the process for many transitions before recording states, reducing the influence of the initial inactive state. Subsequent observations should still be interpreted as dependent observations from one state process.

Recording after one transition targets \(0.20\), not the long-run active proportion \(\frac{2}{3}\). Run the state process through a substantial burn-in before collecting states so the starting condition has much less influence.