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Estimate probabilities by simulation

Learning goals

  • Designs a simulation that represents the random process.
  • Generates or interprets repeated trial results.
  • Computes experimental probabilities from relative frequencies.
  • Compares experimental results with theoretical expectations.
  • Explains how increasing the number of trials affects variability.

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Estimate probabilities by simulation 1
A simulation uses one random digit from \(0\) through \(9\), with all ten digits equally likely. An event should occur with probability \(0.40\). The simulation records the event when the digit is \(0,1,2,\) or \(3\). Is this mapping valid? Explain briefly.

Hints

- 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.

Solution

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.

Answer

Yes. Four of the ten equally likely digits represent the event, so its simulated probability is \(\frac{4}{10}=0.40\).
Estimate probabilities by simulation 2
A simulation of \(20\) repeated trials records \(7\) successes. Use the simulation to estimate the probability of success.

Hints

- 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.

Solution

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

Answer

The estimated probability of success is \(0.35\), or \(35\%\).
Estimate probabilities by simulation 3
Noah wants to estimate how often **both** players make their penalty kicks. In his model, each player has a \(50\%\) chance of scoring, and one kick does not affect the other. He flips two fair coins: heads means a goal and tails means a miss. Does one coin flip or one pair of flips count as one simulation trial? What coin result counts as a successful trial?

Hints

- A simulated trial must represent the whole event being studied. - What does each coin stand for in Noah's model? - Match the success condition to the coin outcomes.

Solution

1. The event involves both kicks, so one simulated trial uses a pair of coin flips. 2. A success is the outcome heads followed by heads, when both kicks are made.

Answer

One trial is a pair of coin flips. A success is two heads.
Estimate probabilities by simulation 4
A ferry company simulates \(20\) morning crossings. The ferry arrives late in \(7\) trials. Estimate the chance that the ferry arrives late on a morning crossing.

Hints

- Does the question ask about one particular crossing or the overall pattern in the trials? - Which of the trial results match the event you are estimating? - How could you express the share as a number between \(0\) and \(1\)?

Solution

1. A trial counts as a success for this event when the ferry is late; this occurred \(7\) times. 2. Divide by all \(20\) trials: \(\frac{7}{20}=0.35\).

Answer

The estimated probability is \(0.35\), or \(35\%\).
Estimate probabilities by simulation 5
A game has a \(50\%\) chance of awarding a ticket. A class proposes using one spin of the wheel for each game. A spin counts as a ticket win when it lands on a sector marked T. Does the wheel model the ticket chance correctly? Explain.
Figure for problem 563295

Hints

- Which features of the sectors matter besides the letters printed on them? - Would counting T labels be enough if the sectors had different sizes? - What share of the entire wheel gives a ticket win?

Solution

1. The wheel has \(6\) equally sized sectors, and \(3\) are marked T. 2. The chance of T is \(\frac{3}{6}=\frac{1}{2}=50\%\), so one spin correctly models a ticket win.

Answer

Yes. Three of the six equally likely sectors are T, giving a \(50\%\) chance.
Estimate probabilities by simulation 6
A finite population has the values \(4,7,9,12,15,18,22,26,31\). A simple random sample of \(5\) distinct values is selected. Describe a simulation to estimate the probability that the sample median exceeds \(15\). State exactly what must be generated and how a trial is classified.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 7
A trial consists of drawing \(3\) tokens without replacement from the urn shown. Describe a simulation that estimates the probability of drawing exactly \(2\) blue tokens. State what must be reset between trials.
Figure for problem 547207

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 8
Leila estimates a probability by running \(100\) simulated trials at a time. After each batch, she calculates the success percentage so far. She stops as soon as the percentage is above \(60\%\) and reports that result. Why might this give a result that is too high? Describe a fair plan for deciding how many trials to run.

Hints

- Would Leila stop after the same number of trials if the early results were different? - Which results get more chances to change before she reports them? - How might two groups comparing results avoid choosing their endpoints differently?

Solution

1. Leila stops after a high estimate but keeps going after a low one. This makes high results more likely to be the ones she reports. 2. A fair plan is to choose the total number of trials before starting, such as \(1000\), then run all of them. 3. Report the total number of successes divided by that fixed trial count, even if the result is below \(60\%\).

Answer

The rule favors unusually high estimates. Choose the number of trials in advance, run them all, and report the success fraction without waiting for a preferred result.
Estimate probabilities by simulation 9
Design a simulation to estimate the probability that at least two of \(8\) randomly selected people share a birth month. Assume all \(12\) months are equally likely and ignore seasonal effects. Specify one trial and the event recorded.

Hints

- 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.

Solution

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.

Answer

Generate \(8\) month labels per trial and record success when a repeated label occurs. The success proportion estimates the probability.
Estimate probabilities by simulation 10
A simulation repeats a trial until the first success or until \(6\) attempts have occurred. In \(2000\) repetitions, the first success occurred on attempts \(1\) through \(6\) with counts \(390,310,252,208,158,130\), and no success occurred in \(552\) repetitions. Estimate the probability that the first success occurs by attempt \(3\), and the probability of no success in \(6\) attempts.

Hints

- 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.

Solution

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 \(\frac{552}{2000}=0.276\).

Answer

The estimates are \(0.476\) for first success by attempt \(3\) and \(0.276\) for no success in \(6\) attempts.
Estimate probabilities by simulation 11
A fair coin is flipped \(12\) times. A simulation records the length of the longest consecutive run of heads in each trial. Describe how one trial should be processed. In \(10{,}000\) trials, the longest run was at least \(4\) heads in \(3118\) trials. Estimate the probability and explain why recording only the total number of heads would be insufficient.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 12
A cereal box contains one of four equally likely puzzle types. Design a simulation to estimate the probability that all four types are collected within \(7\) boxes. State why recording only the number of boxes is not enough.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 13
A simulation program runs \(1000\) trials, but every trial produces exactly the same five pseudorandom numbers. The code calls `set_seed(31415)` at the start of each trial. Diagnose what the seeding instruction is doing. Where should the seed be set if the analyst wants one reproducible simulation run with distinct trials?

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 14
Two appointments independently choose start times uniformly between \(1{:}00\) p.m. and \(3{:}00\) p.m. Each appointment lasts \(30\) minutes. Describe a simulation to estimate the probability that the appointments overlap. State the numerical condition for a successful trial.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 15
A planning team considers two plausible models for an event that can occur on each opportunity: one model uses \(p=0.20\) and the other uses \(p=0.30\). Under each model, the team simulates \(1000\) runs of \(5\) independent opportunities and records whether at least one event occurs. The \(p=0.20\) model produced \(671\) successful runs, and the \(p=0.30\) model produced \(834\) successful runs. A colleague says, “The assumed probability changes by only \(0.10\), so the model choice cannot matter much.” Use the simulation output to assess that claim. Also state one design feature that must be held fixed so the comparison measures sensitivity to \(p\) rather than to some other change.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 16
A bookstore uses a fair random digit from \(0\) through \(9\) to simulate one free gift. The intended chances are \(40\%\) for a sticker, \(30\%\) for a pencil, \(20\%\) for a badge, and \(10\%\) for a bookmark. A student assigns digits \(0\)–\(3\) to stickers, \(4\)–\(6\) to pencils, and \(7\)–\(9\) to badges. In \(100\) trials, no bookmarks occur. a) Why is this not a good simulation of the intended gifts? b) Change the digit assignments to simulate all four gifts fairly. Under your corrected model, about how many bookmarks would you expect in \(100\) trials?

Hints

- Which possible digits produce each kind of prize? - How many of the ten digits should each prize receive? - What fraction of a long run of trials should be bookmarks under a correct rule?

Solution

1. The original rule gives no digit to a bookmark, so its simulated probability is \(0\), not \(0.10\). 2. One valid rule assigns \(0\)–\(3\) to stickers, \(4\)–\(6\) to pencils, \(7\)–\(8\) to badges, and \(9\) to bookmarks. 3. The corrected bookmark chance is \(\frac{1}{10}\), so about \(10\) of \(100\) simulated prizes would be bookmarks; the actual number can vary.

Answer

a) No digit produces a bookmark, so the simulation cannot match its \(10\%\) target chance. b) For example, sticker: \(0\)–\(3\); pencil: \(4\)–\(6\); badge: \(7\)–\(8\); bookmark: \(9\). About \(10\) bookmarks in \(100\) trials.
Estimate probabilities by simulation 17
Students simulate a raffle in which a player can try up to five times to win. They record **which try produced the first win**, or “none” if all five tries lost. The results of \(50\) simulated games are: <table> <tr><th>First win</th><th>Try 1</th><th>Try 2</th><th>Try 3</th><th>Try 4</th><th>Try 5</th><th>None</th></tr> <tr><td>Games</td><td>\(12\)</td><td>\(8\)</td><td>\(7\)</td><td>\(6\)</td><td>\(4\)</td><td>\(13\)</td></tr> </table> a) Estimate the probability of winning **by try 3** and **by try 5**. b) If students had recorded only “won” or “lost,” which estimate could they still make, and which would they no longer know?

Hints

- If a player first wins on try 2, would that count as winning by try 3? - Can a game have its first win both on try 3 and on try 5? - What would a simple won/lost label fail to tell you about timing?

Solution

1. There were \(12+8+7=27\) wins by try \(3\), giving \(\frac{27}{50}=0.54\). 2. There were \(27+6+4=37\) wins by try \(5\), giving \(\frac{37}{50}=0.74\). 3. A final won/lost record still separates the \(37\) successes from \(13\) losses by try \(5\), but cannot tell how many wins happened by try \(3\).

Answer

a) By try \(3\): \(0.54\). By try \(5\): \(0.74\). b) The final win rate \(0.74\) could still be found; the by-try-\(3\) estimate could not.
Estimate probabilities by simulation 18
Twelve balls are independently dropped into four equally likely bins. A simulation estimates the probability that no bin receives more than \(4\) balls. State what one trial must generate and what summary it must retain. In \(8000\) trials, \(3126\) met the condition. Estimate the probability.

Hints

- 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.

Solution

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 \(\frac{3126}{8000}=0.39075\).

Answer

The estimated probability is \(0.39075\), or \(39.075\%\).
Estimate probabilities by simulation 19
Four independent runs of the same simulation each use \(100\) trials. The numbers of successes are \(42\), \(51\), \(47\), and \(44\). a) Explain why the four estimated probabilities are not identical even though the simulation procedure is the same. b) Combine all four runs to give one estimated probability of success.

Hints

- 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.

Solution

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 \(\frac{184}{400}=0.46\).

Answer

a) Random variation makes repeated simulation estimates differ even when the procedure is unchanged. b) \(0.46\)
Estimate probabilities by simulation 20
Two valid simulations estimate the same unknown probability. Run A uses \(50\) trials and records \(31\) successes. Run B uses \(2000\) trials and records \(1190\) successes. a) Find the estimated probability from each run. b) If each simulation were repeated many times, which run size would generally produce estimates that vary less from repetition to repetition? Explain without assuming the true probability is known.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 21
To simulate a \(40\%\) chance of rain, Maya generates one random digit from \(0\) through \(9\), with each digit equally likely. Which four digits could she use for rain? Explain why the choice works.

Hints

- Start with the number of possible digits. - How many of those outcomes should mean rain? - Check the fraction represented by your assignment.

Solution

1. There are \(10\) equally likely digits, so each has probability \(\frac{1}{10}\). 2. Assign rain to four digits, such as \(0,1,2,3\). 3. This gives a rain chance of \(\frac{4}{10}=40\%\).

Answer

One choice is \(0,1,2,3\). Any four distinct digits work because four of ten outcomes represent rain.
Estimate probabilities by simulation 22
A simulation models a fair coin. In a run of \(20\) flips, heads occurs \(13\) times. In a separate run of \(100\) flips, heads occurs \(48\) times. Find each estimate of the probability of heads. Which estimate is closer to \(\frac{1}{2}\)? Must either run give exactly \(\frac{1}{2}\)?

Hints

- Divide each heads count by that run's number of flips. - Compare both estimates to the theoretical chance. - What would be different if every group of flips had to contain exactly half heads?

Solution

1. The estimates are \(\frac{13}{20}=0.65\) and \(\frac{48}{100}=0.48\). 2. Their distances from \(0.50\) are \(0.15\) and \(0.02\), so the second run is closer. 3. A simulated frequency can differ from the theoretical probability; neither run must give exactly \(\frac{1}{2}\).

Answer

\(0.65\) and \(0.48\). The \(100\)-flip run is closer. No run is guaranteed to give exactly \(0.50\).
Estimate probabilities by simulation 23
A prize game awards a sticker \(25\%\) of the time. Liam simulates each play by rolling a fair six-sided die. He calls \(1\) a win and \(2,3,4,5,6\) a loss. Explain the problem with this simulation. Describe a fair chance device that would work instead.

Hints

- What fraction of the die's equally likely faces count as a win? - Write the target chance as a simple fraction. - Choose a device with a number of equal outcomes that fits that fraction.

Solution

1. The die assignment makes the win probability \(\frac{1}{6}\), not \(\frac{1}{4}\). 2. A spinner with four equal sectors, one marked win, correctly gives probability \(\frac{1}{4}=25\%\).

Answer

The die simulates a \(\frac{1}{6}\) chance. Use four equally sized spinner sectors with exactly one winning sector.
Estimate probabilities by simulation 24
A class uses the balls in the jar to simulate drawing two raffle tickets without replacement. A red ball represents a winning ticket. a) How many draws make one complete trial? b) If the first ball is red, how many red balls remain for the second draw? c) What should students do with the balls before starting the next trial?
Figure for problem 563299

Hints

- Match one simulated trial to one complete real experiment. - Without replacement, the first ball stays out during the second draw. - What starting conditions must each new trial have?

Solution

1. The real raffle draws two tickets, so one trial needs two draws without replacing the first ball. 2. The jar starts with \(3\) red balls, leaving \(2\) after a red first draw. 3. Both balls must be returned before a new trial to restore the starting conditions.

Answer

a) Two draws. b) \(2\) red balls. c) Return both balls before the next trial.
Estimate probabilities by simulation 25
Kwame wants to simulate a fair six-sided die using random digits from \(0\) to \(9\). His first rule reports die result \(1\) for digits \(0,6\), result \(2\) for \(1,7\), result \(3\) for \(2,8\), result \(4\) for \(3,9\), result \(5\) for \(4\), and result \(6\) for \(5\). a) Is the simulated die fair? Give the chance of each result. b) Give a fair rule that accepts some digits and tries again after the others. c) What fraction of digits would your rule reject in the long run?

Hints

- For each die face, count the digits that produce it. - What must be equal for all six faces to be equally likely? - What should happen when a generated digit is not used in a completed trial?

Solution

1. Four results each have two digits out of ten, giving \(0.20\) each. The remaining two results have one digit each, giving \(0.10\) each. The die is not fair. 2. One fair rule accepts digits \(0\) through \(5\), maps each to a different die face, and draws another digit if \(6,7,8,9\) occurs. 3. Four of ten digits are rejected, so the long-run rejected share is \(4/10=0.40\).

Answer

a) Results \(1,2,3,4\): \(0.20\) each; results \(5,6\): \(0.10\) each. Not fair. b) Assign digits \(0\)–\(5\) to distinct die faces and redraw after \(6\)–\(9\). c) \(0.40\), or \(40\%\).
Estimate probabilities by simulation 26
A programmer has one long sequence of \(1000\) simulated coin flips. To create \(991\) “trials” of \(10\) flips, the programmer uses flips \(1\) through \(10\), then \(2\) through \(11\), then \(3\) through \(12\), and so on. The event is “at least \(7\) heads.” Explain why these trials are not independent and describe a corrected design that uses the same \(1000\) flips.

Hints

- 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.

Solution

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.

Answer

The overlapping windows are dependent because adjacent trials share \(9\) flips. Use \(100\) nonoverlapping blocks of \(10\) flips, or generate independent blocks separately.
Estimate probabilities by simulation 27
Eight teams are placed uniformly at random into a fixed single-elimination bracket. Teams A and B are the two highest-ranked teams. Describe a simulation to estimate the probability that A and B are placed in opposite halves of the bracket, so their first possible meeting is the championship. What feature of the random placement determines success?

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 28
To simulate a voter selecting option A in a mock poll, Yuna uses one random digit from \(0\) through \(9\) and lets \(8\) digits represent the event. The target probability is about \(0.55\). Evaluate the simulation rule and, if needed, propose a better one-digit rule.

Hints

- 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.

Solution

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.

Answer

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 \(0.55\).
Estimate probabilities by simulation 29
A random-number generator returns a real number uniformly from \(0\) up to but not including \(1\). Rafael uses the rule “success if the output is between \(0.00\) and \(0.30\), inclusive.” What probability does the intended interval represent, and why is the word “inclusive” at the upper endpoint irrelevant for this continuous generator but potentially relevant for a generator restricted to hundredths?

Hints

- 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.

Solution

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.

Answer

The continuous probability is \(0.30\). Endpoint inclusion does not matter continuously, but it can add an outcome in a discrete hundredths generator.
Estimate probabilities by simulation 30
A stick of length \(1\) is broken at two independently and uniformly selected points. The three resulting piece lengths form a triangle exactly when the longest piece is shorter than \(\frac12\). Describe a simulation to estimate this probability. Explain why the two break points must be sorted before the three lengths are calculated.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 31
A simulation needs two fair dice to show a sum of at least \(10\). Compare these ways to generate a trial: 1. Roll two dice and start over whenever the sum is less than \(10\). 2. Choose equally likely from only the ordered dice pairs whose sum is at least \(10\). List the possible accepted pairs. Explain why both methods are fair and which method always finishes in one selection.

Hints

- Which ordered results give a sum of at least \(10\)? - Are any of those results more likely than the others when rolling fair dice? - Which method can produce an unsuccessful first attempt?

Solution

1. The accepted ordered pairs are \((4, 6)\), \((5, 5)\), \((6, 4)\), \((5, 6)\), \((6, 5)\), and \((6, 6)\). 2. All \(36\) original ordered dice pairs are equally likely. Keeping only successful pairs makes each of those six equally likely. 3. Choosing equally from the six pairs directly produces the same chance model. 4. The direct method always takes one selection. Repeated dice rolls might need several attempts.

Answer

The accepted pairs are \((4, 6)\), \((5, 5)\), \((6, 4)\), \((5, 6)\), \((6, 5)\), and \((6, 6)\). Both methods select those pairs equally often in the long run. Direct selection always finishes in one step.
Estimate probabilities by simulation 32
A token starts at \(0\). On each step it moves \(+1\) with probability \(0.60\) and \(-1\) with probability \(0.40\). A trial stops when the token first reaches \(+3\) or \(-2\). Describe a simulation that estimates the probability of reaching \(+3\) first. Identify the state information that must be updated after every step.

Hints

- 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.

Solution

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.

Answer

Simulate the running position step by step, stopping at \(+3\) or \(-2\), and record success when \(+3\) is reached first.
Estimate probabilities by simulation 33
The diagram shows the sampling square and hit region for a Monte Carlo area estimate. Generate points uniformly in the square and count a point as a hit when it lies in the shaded region. In \(50{,}000\) simulated points, \(39{,}241\) were hits. Estimate the shaded area and use it to estimate \(\pi\).
Figure for problem 547224

Hints

- 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.

Solution

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 \(4\cdot0.78482=3.13928\).

Answer

Estimated quarter-circle area: \(0.78482\). Estimated \(\pi\): \(3.13928\).
Estimate probabilities by simulation 34
A clinic simulates a screening test. Each trial follows one full path through the probabilities for risk group and test result. a) Describe how to run one trial using two independent, uniformly chosen random numbers from \([0,1)\). State which number intervals give each result. b) The class runs \(200\) trials. Of these, \(36\) end in high-risk positive, \(18\) in low-risk positive, and the rest in negative results. Estimate the share of **positive simulated tests** that came from high-risk cases.
Figure for problem 547226

Hints

- Why does a number in a short interval represent a less common outcome than a number in a long one? - Should the second number use the same cutoff for both risk groups? - For part b), does the question ask about all trials or only the positive ones?

Solution

1. Let the first random number be \(u\). Choose high risk when \(0\le u<0.20\) and low risk when \(0.20\le u<1\). Choose a second independent number \(v\). For high risk, choose positive if \(0\le v<0.90\), otherwise negative. For low risk, choose positive if \(0\le v<0.10\), otherwise negative. That completes one trial. 2. There are \(36+18=54\) positive trials. Of these, \(36\) began in the high-risk group. 3. The requested simulated share is \(\frac{36}{54}=\frac{2}{3}\).

Answer

a) With \(u\in[0,1)\): high risk for \(u<0.20\), low risk otherwise. With a separate \(v\in[0,1)\): positive for \(v<0.90\) if high risk or \(v<0.10\) if low risk; negative otherwise. b) \(\frac{36}{54}=\frac{2}{3}\), or about \(66.7\%\).
Estimate probabilities by simulation 35
Eight people, including Ava and Ben, sit at random in the eight seats around the circle. Rotations count as the same seating arrangement. Use seat \(S_0\) as a fixed reference seat. a) Describe a simulation that counts each circular arrangement only once. b) How would you decide whether Ava and Ben are next to each other in a trial? c) Why does fixing Ava at \(S_0\) not make any arrangement more likely?
Figure for problem 547229

Hints

- How many rotated versions represent the same circular seating? - Which seats are immediately next to \(S_0\) around the circle? - Does choosing one fixed seat for Ava affect the order of the other people relative to her?

Solution

1. Place Ava at \(S_0\) and randomly arrange the other seven people in the remaining seven seats. This counts each arrangement up to rotation exactly once. 2. Ben is next to Ava if he sits at \(S_1\) or \(S_7\). 3. Every circular seating can be rotated in exactly one way to put Ava at \(S_0\). Fixing her position therefore removes duplicate rotations without favoring any arrangement.

Answer

a) Fix Ava at \(S_0\) and randomly arrange the other seven people. b) A trial succeeds if Ben sits at \(S_1\) or \(S_7\). c) Every circular seating has one rotation with Ava in \(S_0\), so none is favored.
Estimate probabilities by simulation 36
A simulation is intended to estimate the probability that a customer eventually makes a purchase. Each simulated customer is followed until the customer either makes a purchase or leaves. However, trials still active after \(30\) simulated minutes are discarded, and these discarded trials are mostly customers who take a long time to decide. Explain how discarding the unfinished trials can bias the estimated eventual-purchase probability. Give a better simulation rule.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 37
A school carnival is comparing two prize-wheel designs. Design One awards a prize on \(00\)–\(39\) and Design Two on \(00\)–\(49\), using equally likely random two-digit numbers \(00\)–\(99\). The students use the **same** number for both designs in each trial. Their first \(12\) numbers are \(05,47,63,39,41,88,12,44,76,22,50,37\). a) Use these trials to estimate the prize probability for each design. b) How many trials give a prize under Design Two but not Design One? What is the estimated difference between their prize probabilities? c) Based on the assigned number ranges, what is the **theoretical** difference between the prize probabilities? Why may it differ from the simulation estimate?

Hints

- Mark a win or loss for each design beside every simulated number. - Compare the two win counts using the same number of paired trials. - How many of the \(100\) possible numbers award a prize under each rule?

Solution

1. Design One wins on \(05,39,12,22,37\), or \(5\) of \(12\) trials, estimating \(\frac{5}{12}\). 2. Design Two also wins on \(47,41,44\), for \(8\) of \(12\) trials and an estimate of \(\frac{8}{12}=\frac{2}{3}\). 3. Three paired trials differ, so the estimated difference is \(\frac{3}{12}=\frac{1}{4}\). 4. The ranges have \(40\) and \(50\) winning numbers out of \(100\), giving a theoretical difference of \(0.50-0.40=0.10\). Twelve random trials need not match the theoretical difference.

Answer

a) Design One: \(\frac{5}{12}\); Design Two: \(\frac{2}{3}\). b) Three trials; estimated difference \(\frac{1}{4}\). c) Theoretical difference \(0.10\). An estimate from only \(12\) trials can vary from that value.
Estimate probabilities by simulation 38
A school's sound system has a microphone and an amplifier. The table gives the chances that each stops working during one assembly: <table> <tr><th></th><th>Amplifier fails</th><th>Amplifier works</th></tr> <tr><td>Microphone fails</td><td>\(0.08\)</td><td>\(0.06\)</td></tr> <tr><td>Microphone works</td><td>\(0.04\)</td><td>\(0.82\)</td></tr> </table> a) Assign pairs of digits from \(00\) through \(99\) to the four outcomes so that one random pair of digits simulates one assembly. Every pair is equally likely. b) What is the probability that the microphone fails? That the amplifier fails? c) Someone proposes simulating the two failures independently, using just the two probabilities from part b). Would this reproduce the chance that both devices fail in the table? Explain.

Hints

- Each pair of digits should stand for exactly one of the four outcomes in the table. - How many of the \(100\) equally likely digit pairs should stand for each outcome? - For each device, which two cells include its failure? - What would the chance of two failures be if they occurred independently? Compare that with the table.

Solution

1. Use \(00\) through \(07\) when both fail, \(08\) through \(13\) when only the microphone fails, \(14\) through \(17\) when only the amplifier fails, and \(18\) through \(99\) when both work. 2. The microphone fails in two cells: \(0.08+0.06=0.14\). The amplifier fails in two cells: \(0.08+0.04=0.12\). 3. If the two failures were independent, their joint probability would be \(0.14\cdot0.12=0.0168\). The table instead gives \(0.08\) for both failing, so simulating failures independently would not match this model.

Answer

a) One valid assignment: \(00\)–\(07\): both fail; \(08\)–\(13\): only microphone fails; \(14\)–\(17\): only amplifier fails; \(18\)–\(99\): both work. b) Microphone: \(0.14\); amplifier: \(0.12\). c) No. Independent failures would give \(0.14\cdot0.12=0.0168\) for both failing, but the table gives \(0.08\).
Estimate probabilities by simulation 39
A ten-day forecast is modeled conditional on exactly four rainy days, with every four-day subset of the ten days equally likely. a) Describe a direct simulation of one valid ten-day weather schedule. b) Define a success check for the event “at least two rainy days are consecutive.” c) Explain why simulating ten independent rain/no-rain outcomes with probability \(0.4\) does not sample from the stated conditional model.

Hints

- 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.

Solution

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.

Answer

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 probabilities by simulation 40
A game repeatedly rolls a fair six-sided die and adds the results. A trial stops when the running total reaches or exceeds \(12\). The event is “the total equals exactly \(12\) at stopping.” Describe a simulation and explain why overshooting \(12\) is a failure rather than another roll.

Hints

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

Solution

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.

Answer

Simulate a running total until it first reaches at least \(12\). Count success only when the stopping total is exactly \(12\).
Estimate probabilities by simulation 41
Five teams play a round-robin tournament, so every unordered pair of teams plays exactly once. Assume every game is fair, has no tie, and its outcome is independent of every other game. a) Describe one simulation trial for the entire tournament. b) State how to determine whether the trial has an undefeated team. c) Explain why independently generating a final win total for each team would not be a valid simulation.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 42
The communications network shown has five links. In each trial, every link independently works with probability \(0.80\). The network works when at least one complete route connects \(S\) to \(T\). Describe how to simulate one trial without double-counting trials in which more than one route works. In \(25{,}000\) trials, \(22{,}790\) networks worked. Estimate the reliability.
Figure for problem 547243

Hints

- 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 \(T\).

Solution

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 \(91.16\%\).

Answer

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 \(91.16\%\).
Estimate probabilities by simulation 43
A greenhouse starts in the Normal temperature state. Each hour its state may change, and the chance of each next state depends on the current state. a) Describe a complete trial to simulate whether it is Warm **after exactly six hours**. b) The class runs \(200\) such trials and the greenhouse finishes Warm in \(42\). Estimate the probability of ending Warm. c) Why would using the same fixed chance of Warm on every step give the wrong simulation?
Figure for problem 547244

Hints

- Which current state determines your next set of chances? - Would recording “Warm” after two hours settle what happens after six? - What should be identical at the start of each new trial?

Solution

1. Start in Normal. At every step, choose the next state using the branches under the current state in the tree. Use that new state for the next step. 2. After six transitions, record whether the final state is Warm. Repeat from Normal for each new trial. 3. The estimate is \(\frac{42}{200}=0.21\). 4. The chance of becoming Warm differs depending on the current state, so a single fixed chance would not follow the model.

Answer

a) Start Normal, make six successive state choices using the current state's branch chances, and check the sixth state. Reset to Normal before the next trial. b) \(0.21\), or \(21\%\). c) The next-state chances depend on whether the greenhouse is currently Normal or Warm.
Estimate probabilities by simulation 44
A supposedly fair digit generator produced these counts in \(10{,}000\) trials: <table> <tr><th>digit</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th><th>\(7\)</th><th>\(8\)</th><th>\(9\)</th></tr> <tr><td>count</td><td>\(1011\)</td><td>\(983\)</td><td>\(1005\)</td><td>\(996\)</td><td>\(1008\)</td><td>\(990\)</td><td>\(1017\)</td><td>\(995\)</td><td>\(1002\)</td><td>\(993\)</td></tr> </table> a) If all digits are equally likely, how many times would you expect each digit in \(10{,}000\) trials? b) Which digits differ most from that expected count, and by how many? c) Are the results close to or far from the fair model? Describe one way to check whether any imbalance keeps appearing.

Hints

- How many equally likely digits share the total number of trials? - Compare the size of each count's difference from the expected value. - How could fresh trials help tell a repeating pattern from ordinary chance variation?

Solution

1. With ten equally likely digits, the expected count is \(10000/10=1000\) per digit. 2. Digit \(1\) occurs \(17\) fewer times than expected, and digit \(6\) occurs \(17\) more times. These are the largest absolute differences. 3. The differences are small compared with \(1000\). Run new sets of \(10{,}000\) digits and check whether particular digits repeatedly differ in the same direction.

Answer

a) \(1000\) per digit. b) Digits \(1\) and \(6\), each by \(17\) (below and above, respectively). c) The counts are close to the fair model. Repeat the run and look for a consistent pattern of unusually high or low counts.
Estimate probabilities by simulation 45
Inés tries to simulate a point uniformly in the disk shown. Inés generates independent \(U,V\sim\text{Uniform}(0,1)\), sets \(R=U\) and \(\Theta=2\pi V\), and uses polar coordinates \((R,\Theta)\). a) Explain why this method places too many points near the center. b) Derive a corrected formula for \(R\). c) State a complete uniform simulation algorithm for the disk.
Figure for problem 547210

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 46
A game-show simulation compares “stay” and “switch” in a three-door setup. One door hides a prize. After the contestant chooses a door, the host—who knows the prize location—must open a different door showing no prize. Describe one valid trial that preserves the host’s rule. Explain why choosing the host’s door uniformly from both unchosen doors without checking for the prize can bias the comparison.

Hints

- 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.

Solution

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.

Answer

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.
Estimate probabilities by simulation 47
A small service desk is simulated minute by minute for \(60\) minutes. At the start of each minute, a customer arrives with probability \(0.35\). If the clerk is busy, the customer joins the end of the line. At the end of each minute, the customer in service finishes with probability \(0.40\). Describe the state information and update order needed for one trial estimating the probability that the line ever reaches \(5\) waiting customers. Explain why simulating only the total numbers of arrivals and completions is not enough.

Hints

- 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?

Solution

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.

Answer

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.

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