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- Identify which simulated trials count as successes. - Estimate probability with successes divided by all completed trials. - Express the result as a decimal or percentage.

1. A success is a trial in which the token reaches \(3\) before \(-2\). 2. The simulation estimate is \(\frac{812}{2000}=0.406\).

The estimated probability is \(0.406\), or \(40.6\%\).

- Count how many equally likely digit outcomes there are altogether. - Match the desired probability to the required number of digit outcomes.

1. The ten digits are equally likely. 2. A probability of \(0.30=\frac{3}{10}\) requires three of the ten digits to represent the event. 3. Choice A assigns exactly three digits, so it is valid.

A. Event occurs for digits \(0,1,2\).

- Match the number of assigned balls to the desired probability. - Rewrite \(\frac{3}{5}\) and \(\frac{2}{5}\) with denominator \(10\). - More than one valid assignment of the numbered balls is possible.

1. A fair coin has two outcomes with probability \(0.5\) each, so assign \(5\) numbered balls to each outcome. For example, \(\{1,2,3,4,5\}\) represents heads and \(\{6,7,8,9,10\}\) represents tails. 2. In experiment 2, \(P(\text{red})=\frac{3}{5}=\frac{6}{10}\) and \(P(\text{blue})=\frac{2}{5}=\frac{4}{10}\). Assign \(6\) numbered balls to red and \(4\) to blue. For example, balls \(1\) through \(6\) represent red, and balls \(7\) through \(10\) represent blue.

a) Example: \(\{1,2,3,4,5\}\) for heads and \(\{6,7,8,9,10\}\) for tails. b) Example: balls \(1\) through \(6\) for red and balls \(7\) through \(10\) for blue.

- Assign the ten equally likely digits in the same proportions as the colors in the bag. - Red can occur exactly once in either of two positions.

1. One valid assignment is red \(\{0,1,2\}\) and blue \(\{3,4,5,6,7,8,9\}\). 2. A favorable ordered pair has its first digit in the red set and its second digit in the blue set, or vice versa.

For example, assign \(\{0,1,2\}\) to red and \(\{3,4,5,6,7,8,9\}\) to blue. The favorable ordered pairs are \((r,b)\) and \((b,r)\), where \(r\in\{0,1,2\}\) and \(b\in\{3,4,5,6,7,8,9\}\).

- Compare the denominators on the second branches of each diagram. - Without replacement, both the total number of balls and one color count change after the first draw.

1. In Simulation A, the second-draw probabilities remain \(\frac{4}{10}\) for red and \(\frac{6}{10}\) for blue after either first result. This models drawing with replacement. 2. In Simulation B, the second draw uses the \(9\) remaining outcomes. After red, the probabilities are \(\frac{3}{9}\) and \(\frac{6}{9}\); after blue, they are \(\frac{4}{9}\) and \(\frac{5}{9}\). 3. Therefore, Simulation B correctly models drawing without replacement.

Simulation B, because its second-draw probabilities use the \(9\) remaining balls and change according to the color drawn first.

- A duplicate day number represents the event of interest. - Divide matching trials by all trials. - Treat a simulation result as an estimate subject to random variation.

1. A repeated integer represents at least two people sharing a simulated birthday. 2. The simulation estimate is \(\frac{5076}{10{,}000}=0.5076\). 3. The result suggests a probability near and slightly above \(0.5\), but one random simulation run does not prove the exact inequality.

The estimated probability is \(0.5076\). The run suggests that the true probability is slightly above \(0.5\), but it does not prove that claim.

- Define one trial so that it matches the two future shots in the question. - A success must satisfy the entire event, not only part of it. - Estimate probability with successful trials divided by all trials.

1. One trial consists of generating two free-throw results and recording the ordered pair. 2. The trial is a success only when both results are M. 3. The estimated probability is \(\frac{49}{100}=0.49\).

a) Generate two free-throw results and record both. b) M, M. c) \(0.49\), or \(49\%\)

- Count how many equally likely outcomes must be designated as success. - Determine whether the target probability can be written as \(\frac{k}{6}\) for an integer \(k\). - Two rolls create \(36\) equally likely ordered pairs.

1. A fair die has \(6\) equally likely outcomes. Designate any \(2\) outcomes, such as \(1\) and \(2\), as success. Then \(P(\text{success})=\frac{2}{6}=\frac{1}{3}\). 2. Any event based on one roll must contain an integer number \(k\) of the \(6\) outcomes, so its probability must be \(\frac{k}{6}\). The equation \(\frac{k}{6}=0.4\) gives \(k=2.4\), not an integer, so exact simulation is impossible. 3. Two rolls produce \(36\) equally likely ordered pairs. Designate \(9\) pairs as success. For example, define success as both rolls being at most \(3\). This gives \(3\cdot3=9\) successful pairs and probability \(\frac{9}{36}=0.25\).

a) Designate two die results, such as \(1\) and \(2\), as success. b) One-roll event probabilities must be multiples of \(\frac{1}{6}\), and \(0.4\) is not such a multiple. c) Roll twice and call it a success when both results are at most \(3\); the probability is \(\frac{9}{36}=0.25\).

- A fair simulation must produce all \(12\) outcomes with probability \(\frac{1}{12}\). - Combine devices whose numbers of equally likely outcomes multiply to \(12\). - Check both the range and the probability distribution.

1. The sum ranges only from \(2\) through \(10\), and the sums are not equally likely. Therefore, it is not a valid simulation. 2. Method 1: Toss a coin and roll a six-sided die. If the coin shows heads, use the die result \(1\) through \(6\). If it shows tails, add \(6\) to obtain \(7\) through \(12\). Each outcome has probability \(\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}\). 3. Method 2: Roll a four-sided die and a six-sided die. Convert the six-sided result to one of three equally likely values by mapping \(1,2\) to \(1\), \(3,4\) to \(2\), and \(5,6\) to \(3\). If the four-sided result is \(T\) and the converted value is \(R\), output \(3(T-1)+R\). The \(4\cdot3=12\) combinations are equally likely.

1. No. The sums range from \(2\) to \(10\) and are not equally likely. 2. Use a coin with a six-sided die, or use a four-sided die with a six-sided die grouped into three equally likely outcomes.

- Use combinations because order does not matter. - A coin and die together can create \(12\) equally likely outcomes. - Reject repeated numbers to model drawing without replacement.

1. The number of combinations is \(\binom{12}{3}=220\). 2. Generate one number from \(1\) through \(12\) by tossing the coin and rolling the die: use the die result after heads and add \(6\) after tails. Repeat until three distinct numbers have been generated. If a number repeats within the same simulated drawing, discard that result and generate another number. Because each remaining number is equally likely at every accepted stage, every three-number set is equally likely. The final three-number set is the simulated lottery outcome.

1. \(220\) 2. Use the coin to choose \(1\)–\(6\) or \(7\)–\(12\), use the die to select within that range, and reject repeats until three distinct numbers are obtained.

- Count the total number of letters and the number of occurrences of each letter. - A sequence of \(n\) letters contains \(n-1\) overlapping adjacent pairs. - Use independence to find the probability of each ordered pair. - Compare each observed pair count with its expected count.

1. The sequence contains \(9\) occurrences of \(A\) and \(11\) occurrences of \(B\). Thus, the relative frequencies are \(\frac{9}{20}=0.45\) for \(A\) and \(\frac{11}{20}=0.55\) for \(B\). 2. Counting the \(19\) adjacent pairs gives counts of \(2\) for \(AA\), \(7\) for \(AB\), \(6\) for \(BA\), and \(4\) for \(BB\). The corresponding relative frequencies are \(\frac{2}{19}\approx 0.105\), \(\frac{7}{19}\approx 0.368\), \(\frac{6}{19}\approx 0.316\), and \(\frac{4}{19}\approx 0.211\). 3. Because consecutive letters are independent and each letter has probability \(0.5\), every ordered pair has probability \(0.5\cdot 0.5=0.25\). The expected count for each pair is \(19\cdot 0.25=4.75\). Compared with this value, \(AA\) and \(BB\) occur less often, while \(AB\) and \(BA\) occur more often.

1. The relative frequencies are \(0.45\) for \(A\) and \(0.55\) for \(B\). 2. The counts are \(2\) for \(AA\), \(7\) for \(AB\), \(6\) for \(BA\), and \(4\) for \(BB\); the relative frequencies are approximately \(0.105\), \(0.368\), \(0.316\), and \(0.211\), respectively. 3. The expected count is \(4.75\) for each pair. The observed counts of \(AA\) and \(BB\) are lower, and those of \(AB\) and \(BA\) are higher.

- Count adjacent pairs by moving one position at a time through the sequence. - For independent fair tosses, determine the probability of each two-toss pair. - Compare expected pair counts with the observed pattern, not just the overall heads-and-tails balance.

1. The sequence contains \(12\) heads and \(12\) tails, so the relative frequency of each outcome is \(\frac{12}{24}=0.5\). 2. There are \(23\) overlapping adjacent pairs. The counts are \(0\) for \(HH\), \(12\) for \(HT\), \(11\) for \(TH\), and \(0\) for \(TT\). 3. For a fair coin, each ordered pair has probability \(0.25\), so the expected count of \(HH\) is \(23\cdot0.25=5.75\), and the expected count of \(TT\) is also \(5.75\). The sequence contains neither pair and alternates perfectly throughout. Although its single-outcome frequencies are balanced, its pair pattern is extremely atypical of independent fair-coin tosses and suggests deliberate alternation rather than random generation.

1. The relative frequencies of heads and tails are both \(0.5\). 2. The counts are \(0\) for \(HH\), \(12\) for \(HT\), \(11\) for \(TH\), and \(0\) for \(TT\). 3. The sequence is extremely atypical of fair, independent coin tosses. The expected counts of \(HH\) and \(TT\) are each \(5.75\), but both observed counts are \(0\).

- Count the recorded successes before interpreting the proportion. - It is often simpler to calculate the complement of the event requested. - Judge a discrepancy in relation to the small number of simulation trials.

1. There are \(6\) simulated boxes with at least one defective component, so the estimate is \(\frac{6}{20}=0.30\). 2. The probability of no defective components is \((0.92)^5\approx 0.6591\). Thus the theoretical probability is \(1-(0.92)^5\approx 0.3409\). 3. The simulation estimate differs by about \(0.041\). With only \(20\) trials, substantial random variation is plausible, so the discrepancy does not by itself indicate an invalid model.

a) \(0.30\). b) \(1-(0.92)^5\approx 0.3409\). c) No. A \(20\)-trial simulation can vary noticeably from the theoretical probability.

- Track both randomness and the fixed group-size condition. - After the first draw, count the remaining labels that match it. - Consider what group totals become possible if the urn composition never changes.

1. The urn has three A labels and three B labels. Drawing without replacement makes each ordering of three A labels and three B labels equally likely and guarantees group sizes of three. 2. After the first draw, two of the remaining five balls match its label, so the probability is \(\frac{2}{5}=0.4\). 3. With replacement, group counts could be unequal because all six draws would remain independent with no constraint on the number of A and B labels.

a) Three labels of each type and drawing without replacement guarantee balanced random groups. b) \(\frac{2}{5}=0.4\). c) Replacement could produce group sizes other than three and three.

- Convert the success count into a relative frequency. - Account for both possible orders of the desired two-spin outcome. - Judge the discrepancy relative to the number of trials rather than requiring exact equality.

1. The simulation estimate is \(\frac{74}{200}=0.37\). 2. One sector is R, so \(P(R)=\frac{1}{4}\) and \(P(\text{not }R)=\frac{3}{4}\). Exactly one R can occur in either order, giving \(2\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{3}{8}=0.375\). 3. The estimate differs from the theoretical value by \(0.005\), which is very small for \(200\) trials. The simulation appears consistent with the model.

a) \(0.37\). b) \(\frac{3}{8}=0.375\). c) The results are very close, so the simulation is reasonable.

- Compare the area of the unit circle with the area of the surrounding square. - Use the simulated fraction inside the circle as an estimate of the area ratio. - Relate that estimated ratio back to \(\pi\).

1. The circle has area \(\pi\) and the square has area \(4\), so the probability that a uniformly generated point is inside the circle is \(\frac{\pi}{4}\). 2. The simulated inside fraction is \(\frac{3930}{5000}=0.786\). 3. Therefore, \(\pi\approx4\cdot0.786=3.144\).

\(\pi\approx3.144\)

- Estimate probability with successes divided by trials. - More independent trials usually reduce simulation variability. - Reduced variability is not the same as a deterministic error bound.

1. Simulation A estimates \(\frac{39}{100}=0.39\). 2. Simulation B estimates \(\frac{4172}{10{,}000}=0.4172\). 3. Simulation B is generally more stable because it uses many more independent trials. 4. A larger trial count reduces random variation but does not guarantee that one realized estimate lies within any specified distance of the true probability.

The estimates are \(0.39\) and \(0.4172\). Simulation B is generally more stable, but \(10{,}000\) trials do not guarantee that its error is less than \(0.001\).

- Use the frequencies as weights when averaging the possible sample means. - Compute the center of the original four-value population separately. - Identify the tallest stack in the displayed distribution for the most likely result.

1. Using the displayed counts, the sampling-distribution mean is \(\frac{2\cdot 1+3\cdot 2+4\cdot 3+5\cdot 4+6\cdot 3+7\cdot 2+8\cdot 1}{16}=5\). 2. The population mean is \(\frac{2+4+6+8}{4}=5\). 3. The two means are equal. Repeated sample means are centered at the population mean for this random sampling process. 4. The value \(5\) has count \(4\), the largest displayed frequency, so it is most likely.

a) \(5\). b) \(5\). c) The sampling distribution of the sample mean is centered at the population mean. d) \(5\).

- Write the joint probability implied by independence. - Use the conditional relationship supplied by the historical data for the alternative calculation. - Compare the two joint probabilities to determine the direction of model error.

1. The independent simulation uses \(0.30\cdot 0.40=0.12\). 2. The historical relationship gives \(P(\text{rain and traffic})=P(\text{rain})P(\text{traffic}\mid\text{rain})=0.30\cdot 0.70=0.21\). 3. The simulation produces both events too rarely by \(0.21-0.12=0.09\). It will underestimate the frequency of severe combined delays because the real events are positively associated.

a) \(0.12\). b) \(0.21\). c) The model underestimates joint severe delays by treating positively associated events as independent.

- Average each resample independently before comparing them. - Consider what would happen if five values were drawn without replacement from a five-value data set. - A simulation based on observed data inherits weaknesses of those observations.

1. The resample means are \(\frac{78}{5}=15.6\) for A, \(\frac{81}{5}=16.2\) for B, and \(\frac{75}{5}=15.0\) for C. 2. Replacement allows each observed value to appear zero, one, or several times, creating new samples of the same size that mimic repeated sampling from an estimated population. 3. The simulation can reproduce only values and patterns represented by the five observations. A small or unrepresentative original sample limits the usefulness of the resampling distribution.

a) A: \(15.6\); B: \(16.2\); C: \(15.0\) minutes. b) Replacement creates varied same-size resamples and allows repeated observations. c) The method cannot repair a small or unrepresentative original sample.

- Use the dot counts and x-values to reconstruct the \(20\) simulated outcomes. - Compare how the points far to the right affect the mean and the median. - Choose the center that is less distorted by the displayed tail.

1. Reading the dot plot gives a total of \(215\) boxes across \(20\) trials, so the mean is \(\frac{215}{20}=10.75\) boxes. The middle two ordered values are \(8\) and \(9\), so the median is \(\frac{8+9}{2}=8.5\) boxes. 2. A few long waits, especially the values near \(22\) and \(31\), pull the mean upward, while the median is less affected. 3. The median of \(8.5\) boxes better describes a typical trial because the simulated waiting-time distribution is right-skewed. 4. More trials would reduce random simulation variability and stabilize estimates of the distribution, center, and tail behavior.

a) Mean \(10.75\) boxes; median \(8.5\) boxes. b) A few long waits pull the mean upward. c) The median better represents a typical right-skewed trial. d) More trials reduce random simulation variability.

- Check each random-digit set against the required component probability. - Translate the system rule into simpler component events, using a complement where helpful. - Compare the empirical proportion with the calculated model value.

1. A random digit has ten equally likely outcomes. One failure digit gives probability \(0.10\), and two failure digits give probability \(0.20\). 2. A works with probability \(0.90\). At least one of B or C works with probability \(1-0.20\cdot 0.20=0.96\). By independence, system reliability is \(0.90\cdot 0.96=0.864\). 3. The simulation estimate is \(\frac{4318}{5000}=0.8636\), only \(0.0004\) below the theoretical value, so the results are highly consistent.

a) The digit sets contain the required fractions of ten equally likely outcomes. b) \(0.864\). c) The simulation estimate is \(0.8636\), extremely close to theory.

- Use the shuffled outcomes at least as extreme as the observed result as the numerator. - Interpret the resulting proportion under the no-effect model. - Phrase the conclusion about average outcomes rather than every individual.

1. The estimated tail probability is \(\frac{28}{1000}=0.028\). 2. A result this extreme occurs in only about \(2.8\%\) of no-effect randomizations, so it is unusual under the no-effect model. 3. The simulation provides evidence that the study methods differ in mean effect for the experimental setting. It does not show that every individual responds in the same way.

a) \(0.028\), or \(2.8\%\) b) Unusual under the no-effect model. c) There is evidence of a difference in average outcomes between the methods, not a guarantee for every student.

- Relate the center of repeated estimates to the fixed value used to generate the samples. - Interpret spread across repetitions, not spread among individuals in one sample. - Compare the two sample sizes through their square roots.

1. Random sample proportions vary around the true population proportion, so the simulated center \(0.598\) is close to \(0.60\). 2. Across repeated samples of size \(50\), a sample proportion typically differs from the population proportion by about \(0.069\). 3. Increasing size from \(50\) to \(200\) multiplies \(n\) by \(4\), so the standard deviation is divided by \(\sqrt{4}=2\), giving about \(0.0345\). 4. A rough \(95\%\) range is \(0.60\pm2\cdot0.0345\), or approximately \([0.531,0.669]\).

a) Repeated random-sample proportions are centered near the population proportion. b) A size-\(50\) sample proportion typically varies by about \(0.069\) from \(0.60\). c) About \(0.0345\). d) Approximately \([0.531,0.669]\).

- Read the probability of Route C from the spinner before considering two dispatches. - A simulation estimate comes from successful trials divided by all trials. - Compare the observed count with the count predicted by the theoretical probability.

1. From the spinner, \(P(C)=0.20\). For two independent dispatches, the theoretical success probability is \((0.20)^2=0.04\). 2. The simulation estimate is \(\frac{13}{300}\approx0.0433\). 3. The theoretical expected count is \(300\cdot0.04=12\). The observed count \(13\) is close and is consistent with random variation. 4. Three equal sectors would assign probability \(\frac13\) to every route, which does not match the displayed route probabilities.

a) \(0.04\) b) Approximately \(0.0433\) c) Expected \(12\), observed \(13\); the results are consistent. d) Equal sectors would assign the wrong route probabilities.

- Ask what information the seed determines. - Repeating the same outcome sequence does not increase the number of independent trials. - Reproducibility requires recording seeds, not reusing one sequence as new data.

1. The seed determines the pseudorandom sequence, so using the same seed duplicates the same \(500\) outcomes. 2. Averaging duplicate runs does not create \(1000\) independent trials or add new simulation information. 3. To obtain two reproducible runs, use two distinct recorded seeds that initialize separate pseudorandom streams. The two runs should produce different sequences while the recorded seeds make each run reproducible.

The two runs duplicate the same \(500\) outcomes, so they are not a \(1000\)-trial simulation. Use two distinct recorded seeds so the runs produce different reproducible sequences.

- A relative frequency of zero is not automatically the same as a model probability of zero. - Use the complement probability repeatedly across independent trials. - Judge surprise from the calculated probability of the observed outcome.

1. No. A positive-probability event can fail to appear in a finite simulation. 2. The probability of no event on one trial is \(0.998\), so the probability of zero events in \(500\) trials is \((0.998)^{500}\approx 0.3675\). 3. A probability of about \(36.8\%\) is not small, so zero observed events is not surprising under the model. 4. Rare events produce few successes, so their relative-frequency estimates are unstable unless the trial count is very large.

a) No. b) \((0.998)^{500}\approx 0.3675\). c) Not surprising; the no-event outcome has probability about \(36.8\%\). d) More trials are needed to accumulate enough rare successes for a stable estimate.

- Convert the win and loss counts into a total net payoff before averaging. - Count the equally likely two-die outcomes producing the winning sum. - Interpret expected value as a long-run center rather than a promise for one session.

1. The total simulated payoff is \(198\cdot 5-1002\cdot 1=-12\) dollars, so the average is \(\frac{-12}{1200}=-\$0.01\) per play. 2. A sum of \(7\) has probability \(\frac{6}{36}=\frac{1}{6}\). The expected payoff is \(5\cdot\frac{1}{6}-1\cdot\frac{5}{6}=0\). 3. The simulated average of \(-\$0.01\) is close to the theoretical value \(\$0\), so it is consistent with fairness. 4. Individual outcomes and short runs vary substantially; expected value describes a long-run average, not a guaranteed session result.

a) \(-\$0.01\) per play. b) \(\$0\) per play. c) Yes; the simulation is close to the theoretical fair value. d) Short-run random variation can still produce substantial gains or losses.

- Use signals divided by all correct-process simulations. - Multiply the daily probability by \(300\) for an expected count. - False-alarm performance and bad-process detection require different simulated conditions.

1. The estimated false-alarm probability is \(\frac{62}{5000}=0.0124=1.24\%\). 2. Over \(300\) correct-process days, the expected number is \(300\cdot0.0124=3.72\), or about \(4\) false alarms. 3. To evaluate detection, engineers must also simulate one or more genuinely out-of-control processes and record how often the rule signals.

The estimated false-alarm probability is \(0.0124\), or \(1.24\%\), giving about \(3.72\) false alarms in \(300\) correct-process days. A separate simulation of out-of-control processes is needed to estimate detection rates.

- Turn the number of successful trials into a relative frequency. - Use the complete set of equally likely sequences for the exact comparison. - Think about every possible starting position of the consecutive pattern.

1. The simulated estimate is \(\frac{304}{1000}=0.304\). 2. The exact probability is \(\frac{1224}{4096}\approx0.2988\). 3. The simulation estimate differs by \(0.304-0.2988\approx0.0052\), which is reasonable for \(1000\) trials. 4. A run of four heads can begin in any of nine positions, and sequences can contain overlapping runs, so restricting attention to the first four flips misses valid outcomes.

a) \(0.304\). b) \(\frac{1224}{4096}\approx0.2988\); the estimate is about \(0.0052\) higher. c) A qualifying run may start later than the first flip, and runs can overlap.

- Compare the widths and locations of the two boxes rather than relying only on their labels. - The interquartile range is represented by the width of the box itself. - Separate what these bootstrap results show for this data set from a universal claim about means and medians.

1. The median bootstrap distribution has a much narrower box and is concentrated near the center of the original non-outlying values, while the mean bootstrap distribution is much more spread out toward larger values. Therefore, the median appears more resistant to the unusual value. 2. The mean bootstrap distribution has the larger interquartile range because its box is visibly much wider than the median's box. 3. The comparison concerns this data pattern and this purpose. Other distributions or statistical goals can make the mean an appropriate or preferable summary.

a) The median; its bootstrap boxplot is much more tightly concentrated and is less pulled toward large values. b) The mean bootstrap distribution; its box is much wider, so its interquartile range is larger. c) The result concerns this data pattern and purpose, not every possible distribution or statistical goal.

- Use the signs in the observed data when finding the original average change. - The relevant simulated fraction uses only outcomes at least as favorable as the observed result. - Keep evidence about an effect separate from claims about a larger population.

1. The observed total change is \(4+3+3+2+2+1-1-1=13\), so the mean change is \(\frac{13}{8}=1.625\) minutes. 2. The randomization probability is \(\frac{9}{256}\approx0.0352\). 3. A result this favorable is uncommon under the no-systematic-improvement model, which supports evidence of a positive checklist effect for these technicians. 4. Without random sampling from a broader population, the result cannot automatically be generalized to all technicians.

a) \(1.625\) minutes. b) \(\frac{9}{256}\approx0.0352\). c) The simulation gives evidence of improvement for the studied technicians, but broader generalization requires an appropriate sampling design.

- Give each outcome a number of equally likely two-digit results proportional to its probability. - An item can be accepted along either of two nonoverlapping paths. - Compare the simulated relative frequency with the probability from the full process.

1. For the first inspection, assign \(00\)–\(77\) to pass and \(78\)–\(99\) to fail. Use a second independent two-digit number only after a first failure, assigning \(00\)–\(64\) to pass after rework and \(65\)–\(99\) to fail. 2. The eventual acceptance probability is \(0.78+0.22\cdot 0.65=0.923\). 3. The simulated estimate is \(\frac{1837}{2000}=0.9185\). 4. The estimate is \(0.923-0.9185=0.0045\) below the model probability.

a) Use \(00\)–\(77\) for an initial pass; after an initial failure, use \(00\)–\(64\) for a rework pass. b) \(0.923\). c) The estimate is \(0.9185\), which is \(0.0045\) below the model probability.

- Use the proportion of simulated assignments meeting the stated condition. - Compare the estimate and exact value by their absolute difference. - Distinguish randomizing identities from randomizing group sizes.

1. The simulated probability is \(\frac{267}{1000}=0.267\). 2. The absolute difference from the exact probability is \(|0.267-0.2632|=0.0038\), so the result is consistent with ordinary simulation variability. 3. Drawing from ten cards of each type still randomizes which participants receive each treatment, but it forces equal group sizes. 4. Under the urn procedure, the probability of a group-size difference of at least \(6\) is \(0\).

a) \(0.267\). b) Yes. The difference is \(0.0038\). c) The urn method preserves random assignment while guaranteeing \(10\) participants per group, so the imbalance probability becomes \(0\).

- Match what happens to the first selected object in the real process. - Count which unordered pairs satisfy the color condition. - Changing whether an object is returned changes the second-draw probabilities.

1. Algorithm B matches drawing without replacement because the first token is unavailable for the second draw. 2. There are \(\binom{6}{2}=15\) unordered pairs. Only the red pair and the blue pair match, so the actual probability is \(\frac{2}{15}\approx 0.1333\). 3. With replacement, the probability is \(2\cdot\left(\frac{2}{6}\right)^2+2\cdot\left(\frac{1}{6}\right)^2=\frac{5}{18}\approx 0.2778\). 4. Algorithm A overestimates the event by \(\frac{5}{18}-\frac{2}{15}=\frac{13}{90}\approx 0.1444\).

a) Algorithm B. b) \(\frac{2}{15}\approx0.1333\). c) Algorithm A gives \(\frac{5}{18}\approx0.2778\), an overestimate of about \(0.1444\).

- Estimate from the fraction of trials containing at least one head. - Use the complement of five tails for the exact probability. - Trial length is not the denominator for the probability of a trial-level event.

1. The simulation estimate is \(\frac{970}{1000}=0.970\). 2. The exact probability is \(1-\left(\frac{1}{2}\right)^5=\frac{31}{32}=0.96875\). 3. The simulation estimate is \(0.00125\) higher. Event probability is estimated with successful trials divided by all trials, not with the number of flips used.

The simulation estimate is \(0.970\); the exact probability is \(\frac{31}{32}=0.96875\). The estimate is \(0.00125\) high, and its denominator is the \(1000\) completed trials.

- Count how many equally likely digits are assigned to each outcome. - Compare the simulated distribution category by category with the target distribution. - A larger set of equally likely labels can represent hundredths exactly.

1. The one-digit assignment produces probabilities \(0.10\), \(0.30\), and \(0.60\). 2. It understates early delivery by \(0.15-0.10=0.05\) and overstates on-time delivery by \(0.30-0.25=0.05\); the late probability is correct. 3. An exact assignment is \(00\)–\(14\) for early, \(15\)–\(39\) for on time, and \(40\)–\(99\) for late.

a) Early: \(0.10\); on time: \(0.30\); late: \(0.60\). b) Early is \(0.05\) too low and on time is \(0.05\) too high. c) One valid assignment is \(00\)–\(14\), \(15\)–\(39\), and \(40\)–\(99\), respectively.

- Use each possible roll count as a value and its frequency as a weight. - Locate where the cumulative count first passes the halfway position. - A random process contains more information than its average step size.

1. The weighted total is \(2\cdot 75+3\cdot 344+4\cdot 357+5\cdot 163+6\cdot 49+7\cdot 10+8\cdot 2=3805\), so the estimated mean is \(\frac{3805}{1000}=3.805\) rolls. 2. The cumulative frequency through \(3\) rolls is \(419\), and through \(4\) rolls it is \(776\), so the median is \(4\) rolls. 3. The estimated probability of ending within \(3\) rolls is \(\frac{75+344}{1000}=0.419\). 4. Using \(3.5\) every time removes roll-to-roll variability and cannot model the distribution of stopping times.

a) \(3.805\) rolls. b) \(4\) rolls. c) \(0.419\). d) Constant rolls of \(3.5\) preserve only the single-roll mean, not the random paths or stopping-time distribution.

- Check each two-digit value against the assigned success interval. - Look for information reused from one proposed trial to the next. - Partition the digit string into disjoint blocks starting at the first digit.

1. Among the overlapping pairs, \(14\) and \(15\) are in \(00\)–\(24\), so there are \(2\) successes. 2. Consecutive overlapping pairs share one digit; for example, \(31\) and \(14\) both use the digit \(1\). Therefore adjacent trials are dependent. 3. The nonoverlapping pairs are \(31,41,59,26,53,58,97,93\). 4. None of these eight pairs is in \(00\)–\(24\), so the success count is \(0\).

a) \(2\) successes. b) Adjacent pairs share a digit, so the simulated trials are not independent. c) \(31,41,59,26,53,58,97,93\); \(0\) successes.

- Check whether each recorded outcome receives the same number of accepted die faces. - A rejected roll is not itself a recorded outcome. - Multiply the number of rolls by the probability that a roll is accepted.

1. Procedure 1 is correct because, conditional on a roll in \(1\)–\(5\), all five accepted values are equally likely. 2. Procedure 2 gives probabilities \(\frac{1}{6}\) for A, B, C, and D, and \(\frac{2}{6}=\frac{1}{3}\) for E. 3. Each roll is accepted with probability \(\frac{5}{6}\), so \(600\cdot\frac{5}{6}=500\) recorded outcomes are expected.

a) Procedure 1. b) A–D each have probability \(\frac{1}{6}\); E has probability \(\frac{1}{3}\). c) \(500\) recorded outcomes.

- Use one random decision for each matched unit rather than for each individual. - Track both the overall treatment counts and the composition within every pair. - Consider what independent decisions could do to two partners from the same pair.

1. For each of the six pairs, flip one fair coin. Assign A to the first-listed student after heads and to the second-listed student after tails; assign B to the partner. 2. The procedure guarantees exactly six students per method and exactly one student from each matched pair in each method. 3. Independent flips can produce unequal group sizes and can place both members of a matched pair in the same treatment, weakening the intended control for pretest score.

a) Flip one fair coin independently within each of the six pairs and assign opposite methods to the two partners. b) It guarantees equal treatment sizes and one member of every pair in each treatment. c) Independent student-level flips can unbalance both total group sizes and the distribution of matched pretest levels.

- Interpret the sign of each difference before counting favorable scenarios. - Average the scenario-by-scenario differences, not the two policy totals separately. - Ask what source of random variation is canceled when both policies face the same inputs.

1. The differences sum to \(12\), so the mean difference is \(\frac{12}{10}=1.2\) minutes. 2. A positive difference means Policy X had the longer wait, so Policy Y was shorter in \(7\) scenarios. 3. Using matched random inputs removes variation caused merely by different customer streams, so each difference more directly reflects the policy change.

a) \(1.2\) minutes. b) \(7\) scenarios. c) Matched inputs isolate the policy effect by holding the simulated customer conditions constant within each comparison.

- Use the fraction of simulated records at least as extreme as the observed record. - Judge consistency by how often the model reproduces the observed feature. - Think of a mechanism that could make neighboring days more alike.

1. The simulated tail probability is \(\frac{9}{5000}=0.0018\). 2. A result at least this clustered occurs in only about \(0.18\%\) of simulated records, so the observed pattern is not reasonably consistent with the independent-failure model. 3. A plausible feature to investigate is a persistent condition, such as an unresolved mechanical problem, that raises failure risk on neighboring days.

a) \(0.0018\), or \(0.18\%\). b) No. The observed clustering is extremely unusual under the independence model. c) For example, investigate a persistent equipment condition that carries elevated risk from one day to the next.

- Each simulated arrangement should be a uniformly random permutation. - Divide successful trials by total trials. - Distinguish long-run stabilization from a guarantee for one finite simulation.

1. In each trial, generate a random permutation of \(1\) through \(100\). Check whether any position \(i\) contains the value \(i\), record a success if so, and repeat many times. 2. The relative frequency is \(\frac{3162}{5000}=0.6324\). 3. The estimate differs from \(0.6321\) by only \(0.0003\), so this simulation happened to be very close. The law of large numbers explains why estimates tend to stabilize as the number of trials grows, but it does not guarantee such a small error in every \(5000\)-trial simulation.

a) Repeatedly generate a random permutation and check whether at least one value is in its own position. b) \(0.6324\) c) The estimate is very close to \(0.6321\), but a finite simulation has no guarantee of that exact level of accuracy.

- Distinguish “no success yet” from “success occurred on this attempt.” - Think about what replacing every large outcome by the same smaller boundary value does to an average. - If the true waiting time is still unknown at the cap, what information can the simulation honestly record about that trial?

1. A trial with ten failures has waiting time greater than \(10\), not equal to \(10\). The program knows only that the waiting time exceeds \(10\). 2. Replacing every waiting time greater than \(10\) by \(10\) makes large values too small, so the simulated mean is biased downward. 3. For an ordinary Monte Carlo mean, continue each simulated trial until the first success and record that actual attempt number. 4. If a cap must be kept, record a trial that reaches the cap as unresolved or right-censored rather than as \(10\). Such trials cannot be inserted into an ordinary arithmetic mean as if their waiting times were known; a method designed for censored observations would be needed.

a) Ten failures imply a waiting time greater than \(10\), not exactly \(10\). b) The simulated mean is biased downward. c) Continue each trial until success for an ordinary Monte Carlo mean; if a cap is required, mark capped trials as unresolved/right-censored and do not treat them as exact waiting times.

- Track the server's finishing time after every customer. - A customer cannot begin before arriving or before the server becomes free. - Waiting time is the gap between arrival and actual service start.

1. Each start time is the later of the customer's arrival and the previous finish. The start times are \(0, 3, 7, 9, 14\), and the finish times are \(3, 7, 9, 14, 15\). 2. The waiting times are start minus arrival: \(0, 1, 4, 2, 6\) minutes. 3. The mean wait is \(\frac{13}{5}=2.6\) minutes, and the maximum wait is \(6\) minutes. 4. The update rule is \(\text{start}_i=\max(\text{arrival}_i, \text{finish}_{i-1})\), followed by \(\text{finish}_i=\text{start}_i+\text{service}_i\).

a) Start times: \(0, 3, 7, 9, 14\); waiting times: \(0, 1, 4, 2, 6\) minutes. b) Mean wait \(2.6\) minutes; maximum wait \(6\) minutes. c) Use the later of the next arrival and the previous completion as the next start.

- Update the current state after every digit before interpreting the next digit. - The same digit can represent different outcomes under different current states. - A valid simulation must preserve the stated dependence from one day to the next.

1. Starting after dry, digit \(1\) gives rain. After rain, digit \(7\) gives dry. 2. After dry, digit \(4\) gives dry; digit \(0\) then gives rain. 3. After rain, digit \(8\) gives dry; after dry, digit \(3\) gives dry. 4. The six-day sequence is R, D, D, R, D, D, with \(2\) rainy days. 5. A fixed \(0.20\) rule ignores the current state and would fail to model the higher \(0.60\) persistence of rain after a rainy day.

The sequence is R, D, D, R, D, D, so there are \(2\) rainy days. A fixed \(0.20\) rule is invalid because the next-day probability depends on the current weather state.

- Find where the variability term in the numerator is largest. - Translate the desired precision into an inequality and remove the square root carefully. - Use the inverse square-root relationship to compare precision targets.

1. The product \(p(1-p)\) is largest at \(p=0.5\), where it equals \(0.25\). Using this value produces the largest standard deviation for a fixed \(N\). 2. Require \(\sqrt{\frac{0.25}{N}}\le0.01\). Squaring gives \(\frac{0.25}{N}\le0.0001\), so \(N\ge2500\). 3. Standard deviation varies as \(1/\sqrt{N}\). Halving it requires multiplying \(N\) by \(2^2=4\).

a) \(p=0.5\) maximizes \(p(1-p)\), so it gives the largest required \(N\). b) \(N=2500\). c) Increase the trial count by a factor of \(4\).

- Use successful shuffles divided by total simulated shuffles. - Compare the decimal forms before judging agreement. - Ask whether fixing one song's location leaves every other song's possibilities unchanged.

1. The simulated estimate is \(\frac{1841}{5000}=0.3682\). 2. The exact probability is \(\frac{14833}{40320}\approx0.3679\), so the absolute difference is about \(0.0003\). 3. Song positions in a shuffle are dependent because assigning one song to a position changes the positions available to all other songs. 4. The independent calculation gives \(\left(\frac{7}{8}\right)^8\approx0.3436\), which is not the probability for a random permutation.

a) \(0.3682\) b) The exact value is about \(0.3679\); the difference is about \(0.0003\). c) The position events are dependent because a shuffle assigns positions without replacement.

- The complement event may be easier to combine across independent experiments. - Use the probability that one experiment is not falsely flagged. - Distinguish a per-experiment error rate from an error rate for an entire collection.

1. The simulated estimate is \(\frac{496}{1000}=0.496\). 2. The probability of no false flags in ten independent experiments is \((1-0.066)^{10}=0.934^{10}\approx0.5052\). 3. Therefore the probability of at least one false flag is \(1-0.5052\approx0.4948\). 4. Ten opportunities to trigger the rule create many chances for at least one rare event to occur.

a) \(0.496\). b) \(1-0.934^{10}\approx0.4948\). c) Repeating the test creates ten opportunities for a false flag, so the overall probability accumulates.

- Label the original ticket as generation \(0\), then process the queue one ticket at a time. - Count tickets when they are created, including tickets that the cap prevents from being processed. - Compare the capped process with what could happen if generation-\(3\) tickets were allowed to continue.

1. The generation-\(0\) ticket uses digit \(9\) and creates two generation-\(1\) tickets. The first uses \(7\) and closes. The second uses \(8\) and creates two generation-\(2\) tickets. 2. The first generation-\(2\) ticket uses \(8\) and creates two generation-\(3\) tickets. The second uses \(1\) and closes. The generation-\(3\) tickets are counted but not processed. 3. The total number of tickets is \(1+2+2+2=7\). The greatest generation reached is generation \(3\), and \(2\) active tickets remain when the cap stops the simulation. 4. The cap truncates paths that would continue beyond generation \(3\), so it can lower total ticket counts and limit the simulated process depth relative to the uncapped process.

a) The process creates \(7\) tickets in total. b) Generation \(3\), with \(2\) active generation-\(3\) tickets left unprocessed. c) The cap truncates continuing branches and therefore changes the distribution being estimated.

- Identify which random condition is shared by both trains. - Find the two-delay probability separately within each weather condition. - Compare conditional independence with what happens if the shared weather state is removed.

1. Algorithm A is correct because both trains share the same daily weather state. 2. Under good weather, both are delayed with probability \((0.10)^2=0.01\). Under bad weather, both are delayed with probability \((0.60)^2=0.36\). 3. The model probability is \(0.80\cdot0.01+0.20\cdot0.36=0.08\). 4. For either train, the marginal delay probability is \(0.80\cdot0.10+0.20\cdot0.60=0.20\). Algorithm B makes the train outcomes independent, so it gives \((0.20)^2=0.04\). 5. Algorithm B removes the positive dependence created by shared weather and understates simultaneous delays.

a) Algorithm A. b) \(0.08\) c) Algorithm B gives \(0.04\) because it incorrectly gives the trains separate weather conditions and removes their shared dependence.