Experimental probability and simulations

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.

Difficulty: 1 – 5

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Experimental probability and simulations 1

In \(20\) trials of a game, the player wins \(8\) times. What is the experimental probability of winning?

Hints

- Use the observed number of wins, not a theoretical model. - Divide successful trials by total trials.

Solution

1. Experimental probability is the number of observed wins divided by the total number of trials. 2. \(\frac{8}{20}=\frac{2}{5}=0.4\).

Answer

\(\frac{2}{5}=0.4\), or \(40\%\)

Experimental probability and simulations 2

Lena made \(9\) out of \(20\) free throws in practice. What was her experimental probability of making a free throw? Use it to estimate the chance she makes her next one.

Hints

- Which attempts count as successes? - Compare the number of successes with the total number of attempts.

Solution

1. In the practice results, \(9\) of \(20\) attempts went in. 2. Her experimental probability is \(\frac{9}{20}\). 3. This suggests an estimated chance of \(\frac{9}{20}\) on her next attempt, but it does not guarantee that result.

Answer

Experimental probability: \(\frac{9}{20}\). Estimated chance for the next free throw: \(\frac{9}{20}\).

Experimental probability and simulations 3

Maya plays eight rounds of a ring-toss game. W means a win and L means a loss: W, L, W, W, L, L, W, L What fraction of the games were wins?

Hints

- Count every W in the recorded sequence. - How many games were played in all?

Solution

1. Four results are W. 2. There were eight games, so the fraction of wins is \(\frac{4}{8}=\frac{1}{2}\).

Answer

\(\frac{1}{2}\).

Experimental probability and simulations 4

A fair coin is flipped 10 times, and it lands on heads 7 times. What is the experimental probability of heads? What is the theoretical probability of heads?

Hints

- Use the observed flips for one probability. - Use the equally likely sides of the fair coin for the other. - Does a small set of flips have to give an equal number of heads and tails?

Solution

1. Heads occurred 7 times in 10 flips, so the experimental probability is \(\frac{7}{10}\). 2. A fair coin has two equally likely sides, so the theoretical probability of heads is \(\frac{1}{2}\). 3. The probabilities differ because the results of a small number of flips do not have to match the theoretical probability.

Answer

Experimental: \(\frac{7}{10}\). Theoretical: \(\frac{1}{2}\).

Experimental probability and simulations 5

Two classes each flip a fair coin 10 times. In one class, heads appears 3 times. In the other, heads appears 8 times. Must one class have used an unfair coin? Explain.

Hints

- Does a fair coin have to land on heads exactly half the time in every small set of flips? - What might change if the classes repeat the experiment?

Solution

1. The fractions of heads are \(\frac{3}{10}\) and \(\frac{8}{10}\). 2. A fair coin can produce different numbers of heads in different sets of 10 flips. 3. Neither result alone proves that a coin is unfair.

Answer

No. Different sets of flips can have different results even when both coins are fair.

Experimental probability and simulations 6

A student says, “If I roll a fair die 60 times instead of 6 times, the fraction of sixes is guaranteed to be closer to \(\frac{1}{6}\).” Is the student right? Explain.

Hints

- Think about what “guaranteed” means. - Could a short experiment happen to match the expected fraction exactly?

Solution

1. A larger number of rolls usually gives a more stable estimate of the probability of a six. 2. It does not guarantee a better estimate in each particular experiment.

Answer

No. More rolls usually make the fraction more reliable, but a particular set of 60 rolls is not guaranteed to be closer to \(\frac{1}{6}\).

Experimental probability and simulations 7

A bus arrives late on about 1 in 5 school days. To simulate this, a student puts one slip marked L and four slips marked O in a cup. The student draws a slip, records it, and puts it back before drawing again. What part of the procedure represents one school day? Why is the slip put back?

Hints

- Find the action that is repeated each time a new day is modeled. - What changes in the cup if the slip stays out?

Solution

1. One draw and its recorded result represent one school day. 2. Returning the slip keeps the same five equally likely possibilities for the next day.

Answer

One draw represents one school day. Replacing the slip keeps the same chances for every simulated day.

Experimental probability and simulations 8

Two containers hold different mixtures of blue and white buttons. Container 1: \(30\) blue buttons and \(10\) white buttons Container 2: \(10\) blue buttons and \(30\) white buttons In each experiment, a button was drawn, recorded, and replaced four times. Experiment A: \(1\) blue and \(3\) white Experiment B: \(3\) blue and \(1\) white Which experiment is more likely to have used each container? Explain.

Hints

- Find the theoretical fraction of blue buttons in each container. - Find the experimental fraction of blue draws in each experiment. - Match equal or closest fractions.

Solution

1. In Container 1, the theoretical probability of blue is \(\frac{30}{40} = \frac{3}{4}\). Experiment B has an experimental blue frequency of \(\frac{3}{4}\), so it matches Container 1 more closely. 2. In Container 2, the theoretical probability of blue is \(\frac{10}{40} = \frac{1}{4}\). Experiment A has an experimental blue frequency of \(\frac{1}{4}\), so it matches Container 2 more closely. 3. Therefore, the more likely assignment is Container 1 to Experiment B and Container 2 to Experiment A.

Answer

Container 1 most likely produced Experiment B, and Container 2 most likely produced Experiment A.

Experimental probability and simulations 9

Match each statement with the type of probability it uses, and briefly explain your choice. Statement A: “The prize wheel has \(20\) equal sections, and only one section awards the grand prize, so the probability of winning is \(\frac{1}{20}\).” Statement B: “I tossed this irregular building block \(500\) times. It landed on its narrow side \(142\) times, so I estimate the probability of that result as \(28.4\%\).” Types: - Theoretical probability - Experimental probability

Hints

- Was the probability determined before or after repeated trials? - Are equally likely outcomes being counted? - Did the value come from a mathematical model or from observed data?

Solution

1. Statement A uses theoretical probability because it is based on equally likely outcomes: \(P=\frac{1}{20}\). 2. Statement B uses experimental probability because the estimate is the observed relative frequency \(\frac{142}{500}=0.284\).

Answer

A: Theoretical probability, because the sections are equally likely. B: Experimental probability, because the estimate comes from observed relative frequency.

Experimental probability and simulations 10

For each example, identify the probability approach being used and briefly explain your choice. a) The probability of rolling a prime number on a fair eight-sided die is stated as \(0.5\). b) Statistical data report that the probability of a multiple birth is about \(1.2\%\). c) A meteorologist gives a \(30\%\) chance of rain based on weather data from the past \(50\) years. Approaches: theoretical probability or experimental probability.

Hints

- Ask whether equally likely outcomes are being counted. - Look for values based on observations or historical data. - Decide whether each value comes from a mathematical model or collected data.

Solution

1. Part a) uses theoretical probability. A fair die has equally likely faces, and the prime labels \(2,3,5,7\) give \(P=\frac{4}{8}=0.5\). 2. Part b) uses experimental probability because the value comes from observed statistical data. 3. Part c) uses an experimental probability approach because the estimate is based on past observations.

Answer

a) Theoretical probability, based on equally likely die faces. b) Experimental probability, based on statistical data. c) Experimental probability, based on past weather observations.

Experimental probability and simulations 11

For each statement, identify whether it uses theoretical or experimental probability. Briefly explain your choice. a) Demographic data give the probability that a newborn is male as \(51.4\%\). b) A spinner has \(10\) equal sections labeled \(1\) through \(10\). The probability of landing on a prime number is \(0.4\). c) Insurance records give a \(0.02\%\) probability of a wildlife collision on a particular road per trip.

Hints

- Determine whether the value comes from counting equally likely outcomes. - Look for values based on past observations. - Ask whether physical symmetry makes the outcomes equally likely.

Solution

1. Part a) is experimental probability because the value is based on birth records. 2. Part b) is theoretical probability because the equal sections are equally likely and \(4\) of the \(10\) labels are prime. 3. Part c) is experimental probability because the value comes from past insurance data.

Answer

a) Experimental probability b) Theoretical probability c) Experimental probability

Experimental probability and simulations 12

A computer program randomly arranges the four letters \(A\), \(B\), \(C\), and \(D\), using each letter exactly once. a) Assuming all arrangements are equally likely, find the theoretical probability that \(A\) is first. b) In \(10{,}000\) runs, \(A\) appeared first \(2540\) times. Find the relative frequency and identify the type of probability represented by this value.

Hints

- Compare the one favorable first letter with the four possible first letters. - Relative frequency is occurrences divided by trials. - Distinguish a calculated probability from one estimated using data.

Solution

1. Each of the four letters is equally likely to be first, and only one of them is \(A\). Therefore, \(P(A\text{ first})=\frac{1}{4}=0.25\). This is theoretical probability. 2. The relative frequency is \(\frac{2540}{10000}=0.254\). This is experimental probability because it is based on observed results.

Answer

a) \(0.25\); theoretical probability b) \(0.254\); experimental probability

Experimental probability and simulations 13

A game app uses a hidden rule to decide whether a player wins a star. In \(200\) observed plays, \(150\) won stars. a) Is the \(\frac{150}{200}\) win rate theoretical or experimental probability? b) Why can you not assume that “win” and “lose” each have probability \(\frac{1}{2}\)? c) Suppose each new play has an unchanged \(0.75\) chance of winning, independently of other plays. What tends to happen to the fraction of wins as more games are played?

Hints

- Is the stated rate based on what happened or on a known equally likely model? - Does having two outcomes make their chances equal? - Think about how a steady chance shows up over a large number of independent trials.

Solution

1. The rate \(\frac{150}{200}=0.75\) comes from observed plays, so it is experimental. 2. Two possible results do not have to be equally likely. The game's hidden rules could favor wins. 3. With independent plays at a fixed chance of \(0.75\), the fraction of wins tends to settle near \(0.75\) over many plays, not necessarily in every short run.

Answer

a) Experimental probability. b) Two possible outcomes need not be equally likely. c) The win fraction tends to approach \(0.75\) over many independent plays.

Experimental probability and simulations 14

A weighted two-sided game token is tossed \(1000\) times. It lands star side up \(372\) times and circle side up \(628\) times. a) Explain why an equally likely-outcomes model is not suitable for predicting these probabilities before the experiment. b) Estimate the probability that the token lands circle side up. c) Based on the experiment, estimate how many star-side-up results would occur in \(5000\) tosses.

Hints

- Consider whether the device has physical symmetry. - Use relative frequency to estimate probability. - Multiply the estimated star-side-up probability by \(5000\).

Solution

1. The token is weighted, so there is no reason to assume that its two sides are equally likely to land face up. 2. The estimated probability of landing circle side up is \(\frac{628}{1000}=0.628\). 3. The relative frequency of star-side-up results is \(\frac{372}{1000}=0.372\). The estimated number in \(5000\) tosses is \(0.372\cdot5000=1860\).

Answer

a) The token is weighted, so the two outcomes need not be equally likely. b) \(0.628\) c) Approximately \(1860\) star-side-up results

Experimental probability and simulations 15

A custom deck contains \(16\) cards: the hearts and diamonds with ranks \(7,8,9,10,J,Q,K,A\). One card is drawn at random. a) Assuming all cards are equally likely, find the theoretical probability of drawing an ace. b) In \(400\) draws with replacement, an ace was drawn \(48\) times. Find the relative frequency and compare it with the theoretical probability. c) Does the experimental result provide evidence against the equally likely-card model? Explain using the relationship between relative frequency and probability.

Hints

- Count the aces and the total cards. - Divide the observed ace count by \(400\). - A relative frequency does not have to equal the theoretical probability exactly.

Solution

1. The deck has \(16\) cards and \(2\) aces, so \(P(\text{ace})=\frac{2}{16}=0.125\). 2. The relative frequency is \(\frac{48}{400}=0.12\), which is \(0.005\) below the theoretical probability. 3. The result does not provide meaningful evidence against the model. Random variation can produce a relative frequency slightly different from the theoretical probability, and \(0.12\) is close to \(0.125\).

Answer

a) \(0.125\) b) \(0.12\), which is \(0.005\) below the theoretical value c) No. The small difference is consistent with ordinary random variation.

Experimental probability and simulations 16

Two coin simulations produced the frequencies shown. Which simulation gives more convincing support for the claim that the coin is fair? Justify your answer using relative frequencies.

Hints

- Do not compare only the absolute counts. - Find the proportion of heads in each simulation.

Solution

1. Simulation A: The relative frequency of heads is \(\frac{31}{50} = 0.62\). 2. Simulation B: The relative frequency of heads is \(\frac{2511}{5000} = 0.5022\). 3. The distances from \(0.5\) are \(0.12\) for A and \(0.0022\) for B. 4. Simulation B gives more convincing support because its relative frequency is much closer to \(0.5\) and is based on many more trials.

Answer

Simulation B; its relative frequency of heads, \(0.5022\), is much closer to \(0.5\).

Experimental probability and simulations 17

An event has a theoretical probability of \(0.3\). In \(1000\) simulated trials, the event occurred \(287\) times. Is this outcome reasonable for the probability model? Explain.

Hints

- Calculate the experimental probability from the event count and total trials. - Compare the experimental value with \(0.3\). - A finite simulation does not have to match the theoretical probability exactly.

Solution

1. The experimental probability is \(\frac{287}{1000}=0.287\). 2. Its difference from the theoretical probability is \(0.3-0.287=0.013\). 3. Simulation results vary, and \(0.287\) is close to \(0.3\). The outcome is reasonable for the model.

Answer

Yes. The experimental probability \(0.287\) is close to the theoretical probability \(0.3\).

Experimental probability and simulations 18

A toy car reaches a Y-shaped track junction \(180\) times. It turns left \(108\) times. Use the results to estimate the probability that the car turns left. About how many left turns would you predict in the next \(75\) runs?

Hints

- Compare the number of desired outcomes with the total number of trials. - Use the resulting rate to estimate a count for a new number of trials. - Treat the prediction as approximate rather than guaranteed.

Solution

1. The experimental probability of a left turn is \(\frac{108}{180}=0.6\). 2. The predicted number of left turns is \(0.6\cdot75=45\). 3. The actual number may differ because experimental results vary.

Answer

The estimated probability is \(0.6\), and about \(45\) left turns are predicted.

Experimental probability and simulations 19

A computer simulation models a package passing two inspections. In \(300\) trials, the package passed both inspections \(147\) times. Estimate the probability that a package passes both inspections. Based on the simulation, about how many of \(1000\) packages would pass both?

Hints

- Use successful trials over total trials for the estimate. - Apply the estimated rate to the new total. - Interpret the resulting count as a prediction.

Solution

1. The experimental probability is \(\frac{147}{300}=0.49\). 2. The predicted count is \(0.49\cdot1000=490\). 3. The prediction is approximate because future results can vary.

Answer

The estimated probability is \(0.49\), and about \(490\) of \(1000\) packages are predicted to pass both inspections.

Experimental probability and simulations 20

A computer simulates a carnival game. A probability model says the chance of winning each play is \(\frac{3}{10}\). In \(240\) simulated games, the player wins \(73\) times. Find the experimental probability and its exact difference from the model probability. Do the two values agree closely?

Hints

- Use the number of wins over all simulated games. - Express the model probability with the same denominator as the experimental fraction. - Compare the two numerators to find the difference.

Solution

1. The experimental probability is \(\frac{73}{240}\). 2. The model probability is \(\frac{3}{10}=\frac{72}{240}\). 3. The difference is \(\frac{73}{240}-\frac{72}{240}=\frac{1}{240}\). This is a small difference, so the results agree closely.

Answer

Experimental probability: \(\frac{73}{240}\). Difference: \(\frac{1}{240}\). Yes, the simulation is close to the model probability.

Experimental probability and simulations 21

Two students design simulations for an event with probability \(0.60\). Method A: Generate a random digit \(0\) through \(9\); digits \(0\) through \(5\) mean success. Method B: Roll a fair six-sided number cube; results \(1\), \(2\), or \(3\) mean success. Which method correctly models the event? Explain.

Hints

- Count the equally likely outcomes in each method. - Find the fraction assigned to success. - Compare each fraction with the target probability.

Solution

1. Method A assigns \(6\) of \(10\) equally likely digits to success, so it models \(\frac{6}{10}=0.60\). 2. Method B assigns \(3\) of \(6\) equally likely outcomes to success, so it models \(\frac{3}{6}=0.50\). 3. Therefore, only Method A correctly models the event.

Answer

Method A is correct because it models probability \(0.60\). Method B models probability \(0.50\), not \(0.60\).

Experimental probability and simulations 22

A bag contains \(2\) blue and \(3\) yellow tiles. A student models a game by drawing a tile, replacing it, and mixing the bag before each of three draws. The trial counts as a win if at least two drawn tiles are blue. In \(600\) trials, there were \(221\) wins. Why is replacement important for this model? Find the experimental probability of winning as a fraction, and predict the number of wins in \(1200\) new trials.

Hints

- How would the bag change during a trial if a tile stayed out after the first draw? - What fraction of the recorded trials were wins? - Use that fraction to make a prediction, not a promise, about the new trials.

Solution

1. Replacement keeps the same bag composition for each draw; otherwise later color chances would change. 2. The experimental probability is \(\frac{221}{600}\). 3. Using that rate for \(1200\) trials gives \(\frac{221}{600}\cdot1200=442\) predicted wins. This is not guaranteed.

Answer

Replacement keeps the color chances unchanged from draw to draw. Experimental probability: \(\frac{221}{600}\). Predicted wins: about \(442\), not a guarantee.

Experimental probability and simulations 23

About \(20\%\) of the deliveries at a bike shop arrive after the promised time. Use random digits from 0 through 9, each equally likely, to simulate 30 deliveries. Give a rule for deciding whether a delivery is late, say what one trial is, and explain what you would record.

Hints

- How many of the ten digits should represent a late delivery? - Make each digit stand for one delivery, not several. - Decide what data you need to keep to estimate the late rate.

Solution

1. Label digits 0 and 1 as late; label digits 2 through 9 as on time. That makes the late chance \(\frac{2}{10}=20\%\). 2. Generate one random digit for each delivery, for 30 digits in all. 3. Record the outcome for each digit and count the number of late deliveries. Dividing that count by 30 gives the simulated late fraction.

Answer

Let 0 or 1 mean late and 2–9 mean on time. One random digit represents one delivery. Generate 30 digits, record each result, and count the late deliveries.

Experimental probability and simulations 24

At a school game booth, a visitor draws one of five identical cards. Two cards have a star. A visitor also flips a fair coin, and wins only with a star and heads. Describe a simulation of 40 visitors using the five cards and a coin. Include what one trial is, what must be reset, and what to record.

Hints

- What two things must happen for one visitor to win? - How can each new visitor face the same set of cards? - What single result should you record after both random actions?

Solution

1. For one simulated visitor, draw one of the five cards and flip the coin. 2. Record a win only if the card shows a star and the coin shows heads; otherwise record a loss. 3. Replace the drawn card and mix the cards before the next visitor. 4. Repeat the complete trial 40 times and count the wins.

Answer

For each of 40 trials, draw a card and flip the coin. Count a win only for a star and heads. Replace and mix the card after every trial. Record the number of wins.

Experimental probability and simulations 25

A game is supposed to give a prize on about \(\frac{1}{3}\) of plays. A student simulates it with equally likely random digits from 0 to 9. Digits 0, 1, and 2 mean prize; all other digits mean no prize. Is the simulation set up correctly? If not, give a simple fair-die rule that works.

Hints

- How many of the ten digits mean prize? - Can six equally likely die faces be divided into three equal groups?

Solution

1. Three of the ten digits mean prize, so the simulated probability is \(\frac{3}{10}\), not \(\frac{1}{3}\). 2. A fair six-sided die works if 1 or 2 means prize and 3, 4, 5, or 6 means no prize. 3. Then the chance of prize is \(\frac{2}{6}=\frac{1}{3}\).

Answer

No. The digit rule gives \(\frac{3}{10}\). Instead, roll a fair die and let 1 or 2 mean prize.

Experimental probability and simulations 26

Two independent fair coins are flipped together. A student simulates them by tossing just one coin, then writes the same result for both coins: HH for heads or TT for tails. Why is this simulation unsuitable? Describe a better one.

Hints

- Can two independent coins show different sides? - Which outcomes are impossible under the student's rule? - How can each coin get its own random result?

Solution

1. The student's method produces only HH and TT, so it excludes HT and TH. 2. Independent coins can land differently. 3. Toss two coins, or toss one coin twice, and record both results in order for each trial.

Answer

The simulation leaves out HT and TH. Toss two coins, or toss one coin twice, and record both results.

Experimental probability and simulations 27

A student launches paper airplanes toward a marked target. What is the experimental probability that a flight lands in the target area?

Hints

- Read the count for each result from the bar heights. - What is the total number of trials? - Compare successes with all trials.

Solution

1. The bars show \(8\) flights landing in the target area and \(12\) misses, for \(20\) flights in all. 2. The experimental probability is \(\frac{8}{20}=\frac{2}{5}\).

Answer

\(\frac{2}{5}\).

Experimental probability and simulations 28

Both charts come from flips of a fair coin: a) after \(20\) flips and b) after \(100\) flips. Which experiment's heads proportion is closer to the theoretical value of \(\frac{1}{2}\)? Does that mean every experiment with more flips must be closer? Explain.

Hints

- Read the height of the Heads bar in each chart. - Compare each height with \(0.5\). - A trend over many experiments is not a guarantee for one experiment.

Solution

1. In a), the heads relative frequency is \(0.7\), which is \(0.2\) from \(0.5\). 2. In b), the heads relative frequency is \(0.6\), which is \(0.1\) from \(0.5\). 3. Experiment b) is closer in these results. More flips usually make the proportion more stable, but closeness is not guaranteed for every particular experiment.

Answer

b) is closer to \(\frac{1}{2}\). No; more flips do not guarantee a closer result in every experiment.

Experimental probability and simulations 29

A computer places \(500\) points uniformly at random inside a square. A circle is drawn inside the square so that it touches all four sides. Of the \(500\) points, \(392\) land inside the circle. Use the simulation results to estimate \(\pi\).

Hints

- How is the circle's radius related to the square's side length? - What fraction of the square's area is inside the circle? - How can the fraction of random points inside the circle estimate the area ratio? - How can you solve the area-ratio equation for \(\pi\)?

Solution

1. The experimental proportion of points inside the circle is \(\frac{392}{500}=0.784\). 2. If the square has side length \(s\), then the circle has radius \(\frac{s}{2}\). The area ratio is \(\frac{A_{\text{circle}}}{A_{\text{square}}}=\frac{\pi(\frac{s}{2})^2}{s^2}=\frac{\pi}{4}\). 3. Use the observed proportion to estimate the area ratio: \(\frac{\pi}{4}\approx0.784\). 4. Multiply by \(4\): \(\pi\approx4\cdot0.784=3.136\).

Answer

\(\pi\approx3.136\)

Experimental probability and simulations 30

Three student groups estimate how often the letter S appears in English text. - Group A examines \(50\) letters and finds S \(2\) times. - Group B examines \(500\) letters and finds S \(36\) times. - Group C combines the class results: S appears \(310\) times among \(5000\) letters. a) Find the relative frequency of S for each group. b) Which value is probably the best estimate of the long-run proportion of S in similar English texts? Justify your choice using a statistical principle.

Hints

- Compare the number of letters examined by each group. - Think about what usually happens to relative frequency as the sample size grows. - A larger sample is less affected by a few unusual observations.

Solution

1. a) Group A: \(\frac{2}{50} = 0.04 = 4\%\). Group B: \(\frac{36}{500} = 0.072 = 7.2\%\). Group C: \(\frac{310}{5000} = 0.062 = 6.2\%\). 2. b) Group C provides the best estimate because it uses the largest sample. By the law of large numbers, relative frequencies generally become more stable as the number of observations increases. A larger sample does not guarantee an exact value, but it usually reduces the effect of random variation.

Answer

a) Group A: \(4\%\); Group B: \(7.2\%\); Group C: \(6.2\%\) b) Group C, because its sample is largest and its relative frequency is generally the most stable estimate.

Experimental probability and simulations 31

A computer program counts vowels in a passage containing exactly \(250\) letters: A: \(15\), E: \(32\), I: \(20\), O: \(8\), U: \(12\). a) Find the relative frequency of vowels in the passage. b) Ivana says, “If a reference model gives the letter E a probability of \(12.7\%\), then every \(250\)-letter English passage must contain E exactly \(31.75\) times.” Evaluate the claim using both the calculation and the meaning of a probability model.

Hints

- For part a), think about the fraction of all letters that are vowels. - For part b), ask whether a probability-model expectation must be a possible count in one individual passage. - Distinguish a long-run model prediction from what must happen in every sample.

Solution

1. The total number of vowels is \(15 + 32 + 20 + 8 + 12 = 87\). 2. The relative frequency of vowels is \(\frac{87}{250} = 0.348 = 34.8\%\). 3. The model calculation is \(250 \cdot 0.127 = 31.75\). This is an expected or long-run average count, not a required count in every passage. 4. A letter count must be a whole number, and individual passages vary because of their vocabulary and subject. Therefore, Ivana's claim is false.

Answer

a) \(34.8\%\) b) The claim is false. The value \(31.75\) is a model-based expected count, not a count that every passage must have, and actual letter counts must be whole numbers.

Experimental probability and simulations 32

A computer simulates drawing one ball from a bag containing ten balls numbered \(1\) through \(10\). The ball is replaced after every draw. The table shows how often the number \(7\) was drawn in different numbers of trials. <table> <tr><td>Number of trials \(n\)</td><td>\(100\)</td><td>\(1000\)</td><td>\(10{,}000\)</td><td>\(100{,}000\)</td></tr> <tr><td>Number of 7s</td><td>\(13\)</td><td>\(92\)</td><td>\(1045\)</td><td>\(9982\)</td></tr> </table> a) Find the relative frequency of drawing a \(7\) for each set of trials. b) Find the theoretical probability of drawing a \(7\). c) Compare the relative frequencies with the theoretical probability. Explain how the results illustrate the law of large numbers.

Hints

- Divide the number of 7s by the total number of trials. - All ten numbered balls are equally likely. - Compare each experimental result with \(0.1\).

Solution

1. a) For \(n = 100\), the relative frequency is \(\frac{13}{100} = 0.13\). For \(n = 1000\), the relative frequency is \(\frac{92}{1000} = 0.092\). For \(n = 10{,}000\), the relative frequency is \(\frac{1045}{10000} = 0.1045\). For \(n = 100{,}000\), the relative frequency is \(\frac{9982}{100000} = 0.09982\). 2. b) The ten outcomes are equally likely, so \(P(7) = \frac{1}{10} = 0.1\). 3. c) The absolute differences from \(0.1\) are \(0.03\), \(0.008\), \(0.0045\), and \(0.00018\). In these trial sets, the relative frequency gets closer to \(0.1\) as the number of trials increases. This illustrates the law of large numbers: over many repetitions, experimental probability tends to stabilize near theoretical probability.

Answer

a) \(0.13\), \(0.092\), \(0.1045\), and \(0.09982\) b) \(0.1 = 10\%\) c) As the number of trials increases, the relative frequency shown becomes closer to \(0.1\), illustrating the law of large numbers.

Experimental probability and simulations 33

Two seventh-grade classes investigate whether a coin is fair. Class 7A flips the coin \(250\) times and gets tails \(140\) times. Class 7B flips the coin \(2500\) times and gets tails \(1280\) times. a) Find the relative frequency of tails for each class. b) For a fair coin, the theoretical probability of tails is \(50\%\). Use the law of large numbers to explain which result gives more reliable evidence about whether the coin is fair.

Hints

- Compare the number of flips completed by the two classes. - Consider how sample size affects random variation. - Decide which relative frequency is the more stable estimate of \(0.5\).

Solution

1. a) Class 7A: \(\frac{140}{250} = 0.56\). Class 7B: \(\frac{1280}{2500} = 0.512\). 2. b) Class 7B used ten times as many flips. By the law of large numbers, relative frequency generally becomes more stable around theoretical probability as the number of trials increases. Therefore, the value \(0.512\) from Class 7B is the more reliable estimate. The law does not require the results to move closer to \(0.5\) after every additional flip.

Answer

a) Class 7A: \(0.56 = 56\%\); Class 7B: \(0.512 = 51.2\%\) b) Class 7B gives more reliable evidence because it used many more trials, so random variation has less influence on the relative frequency.

Experimental probability and simulations 34

Two bags contain gold and silver marbles. Bag A contains \(10\) gold marbles and \(90\) silver marbles. Bag B contains \(50\) gold marbles and \(50\) silver marbles. Chinedu repeatedly drew a marble from one bag, recorded its color, and replaced it. In \(10\) draws, Chinedu drew gold \(4\) times and silver \(6\) times. Which bag was Chinedu more likely using? Explain by comparing the theoretical probabilities with the experimental result.

Hints

- Compare the long-run chance of gold from each bag with the observed proportion of gold draws. - Ask which theoretical probability is closer to the experimental result. - Remember that a small sample can support a more-likely conclusion without proving which bag was used.

Solution

1. In Bag A, the theoretical probability of gold is \(\frac{10}{100} = \frac{1}{10}\). 2. In Bag B, the theoretical probability of gold is \(\frac{50}{100} = \frac{1}{2}\). 3. The experimental relative frequency of gold is \(\frac{4}{10} = \frac{2}{5}\). 4. The value \(\frac{2}{5}\) is much closer to \(\frac{1}{2}\) than to \(\frac{1}{10}\), so the results are more consistent with Bag B. This does not prove which bag was used, but Bag B is more likely.

Answer

Bag B is more likely. Its gold-marble probability is \(\frac{1}{2}\), and the experimental result \(\frac{4}{10} = \frac{2}{5}\) is much closer to \(\frac{1}{2}\) than to Bag A's \(\frac{1}{10}\).

Experimental probability and simulations 35

A weighted six-sided die is being tested. Two students discuss the probability of rolling a \(6\). Luke says, “The die has six sides, so every number must have probability \(\frac{1}{6}\).” Sarah says, “I rolled the die \(200\) times and got a \(6\) on \(52\) rolls, so I estimate the probability at about \(26\%\).” 1. Explain the assumption behind Luke’s reasoning and why it may be invalid here. 2. Find the relative frequency Sarah used and name the principle supporting her estimate. 3. Describe how the probability could be estimated more accurately.

Hints

- Ask whether the physical device makes all outcomes equally likely. - Relative frequency is successes divided by trials. - Consider how increasing the number of trials affects random variation.

Solution

1. Luke assumes that all six outcomes are equally likely. Because the die is described as weighted, that symmetry assumption may be false. 2. Sarah’s relative frequency is \(\frac{52}{200}=0.26\). The law of large numbers supports using a long-run relative frequency to estimate an unknown probability. 3. Roll the die many more times under the same conditions. A larger number of trials generally produces a more stable relative-frequency estimate.

Answer

1. Luke assumes equally likely outcomes, which is not justified for a weighted die. 2. The relative frequency is \(0.26\), or \(26\%\); the supporting principle is the law of large numbers. 3. Conduct many more rolls under consistent conditions and recompute the relative frequency.

Experimental probability and simulations 36

A bottle cap is tossed \(500\) times. It lands open side up \(320\) times and top side up \(180\) times. a) Explain why assuming the two outcomes are equally likely is not appropriate. b) Use the experimental results to estimate the probability that the bottle cap lands open side up. c) Explain how relative frequency is connected to probability as the number of tosses becomes very large.

Hints

- Consider whether the object has physical symmetry. - Divide the number of open-side-up results by the total number of tosses. - Think about how random variation changes with many trials.

Solution

1. A bottle cap is physically asymmetric, so there is no reason to assume that its two resting positions are equally likely. 2. The experimental estimate is \(\frac{320}{500}=0.64\). 3. Under repeated independent trials in the same conditions, the relative frequency tends to stabilize near the underlying probability as the number of trials grows. This is the law of large numbers.

Answer

a) A bottle cap is not symmetric, so the two outcomes need not be equally likely. b) \(0.64\), or \(64\%\) c) With many repeated trials, the relative frequency tends to approach the underlying probability.

Experimental probability and simulations 37

Zofia investigates how a thumbtack lands when tossed: on its round head, event \(H\), or on its side, event \(S\). a) Explain why equally likely outcomes cannot be assumed to predict an exact value of \(P(H)\). b) In \(500\) tosses, the thumbtack lands on its head \(185\) times. Use experimental probability to estimate \(P(H)\). c) State the conditions that \(P(H)\) and \(P(S)\) must satisfy to form a valid two-outcome probability model.

Hints

- Equally likely outcomes require an appropriate symmetry argument. - Experimental probability is an observed relative frequency. - Probabilities for all outcomes in the sample space must be nonnegative and sum to \(1\).

Solution

1. The thumbtack is not physically symmetric, so there is no reason to assume that the two outcomes are equally likely. 2. The experimental estimate is \(P(H)\approx\frac{185}{500}=0.37\). 3. A valid model requires \(P(H)\ge 0\), \(P(S)\ge 0\), and \(P(H)+P(S)=1\).

Answer

a) The outcomes cannot be assumed equally likely because the thumbtack is not symmetric. b) \(P(H)\approx 0.37\). c) \(P(H),P(S)\ge 0\) and \(P(H)+P(S)=1\).

Experimental probability and simulations 38

An urn contains \(4\) red balls and \(6\) blue balls that differ only in color. a) Use theoretical probability to find the probability of drawing a red ball. b) In \(50\) trials with replacement, a red ball is drawn \(18\) times. Find the relative frequency and identify the associated probability approach. c) Explain why the result in part b) can differ from the theoretical probability in part a).

Hints

- For equally likely balls, divide favorable outcomes by total outcomes. - Relative frequency is successes divided by trials. - Consider the effect of random variation in a limited sample.

Solution

1. The theoretical probability is \(P(\text{red})=\frac{4}{10}=0.40\). 2. The relative frequency is \(\frac{18}{50}=0.36\), which is an experimental probability estimate. 3. Relative frequencies vary from sample to sample. With more trials, the relative frequency generally becomes more stable around the theoretical probability.

Answer

a) \(P(\text{red})=0.40\) b) The relative frequency is \(0.36\); this is experimental probability. c) Random variation in a finite number of trials can make the experimental result differ from the theoretical value.

Experimental probability and simulations 39

An irregularly shaped object can come to rest in positions \(A\), \(B\), or \(C\). The table shows cumulative results as the number of tosses increases. <table> <thead> <tr> <th>Number of tosses \(n\)</th> <th>Position \(A\)</th> <th>Position \(B\)</th> <th>Position \(C\)</th> </tr> </thead> <tbody> <tr> <td>\(200\)</td> <td>\(82\)</td> <td>\(74\)</td> <td>\(44\)</td> </tr> <tr> <td>\(400\)</td> <td>\(170\)</td> <td>\(142\)</td> <td>\(88\)</td> </tr> <tr> <td>\(600\)</td> <td>\(250\)</td> <td>\(218\)</td> <td>\(132\)</td> </tr> <tr> <td>\(800\)</td> <td>\(336\)</td> <td>\(288\)</td> <td>\(176\)</td> </tr> </tbody> </table> a) Find the relative frequency of each position after \(800\) tosses. b) Based on these data, estimate how many times each position would occur in \(2400\) tosses.

Hints

- Divide each count by the total number of tosses. - Use the largest available trial count for the most stable estimate. - Multiply each estimated probability by \(2400\).

Solution

1. After \(800\) tosses, the relative frequencies are \(\frac{336}{800}=0.42\) for position \(A\), \(\frac{288}{800}=0.36\) for position \(B\), and \(\frac{176}{800}=0.22\) for position \(C\). 2. Using these relative frequencies as estimates, the predicted counts in \(2400\) tosses are \(0.42\cdot2400=1008\) for \(A\), \(0.36\cdot2400=864\) for \(B\), and \(0.22\cdot2400=528\) for \(C\).

Answer

a) Position \(A\): \(0.42\); position \(B\): \(0.36\); position \(C\): \(0.22\) b) Approximately \(1008\) times in position \(A\), \(864\) times in position \(B\), and \(528\) times in position \(C\)

Experimental probability and simulations 40

A large transportation study in a metropolitan area found that \(18\%\) of households have no car, \(54\%\) have exactly one car, and \(28\%\) have two or more cars. a) Describe the random selection process and model assumption needed for the statement “The probability that a randomly selected household has no car is approximately \(0.18\)” to be meaningful. b) In the context of the law of large numbers, distinguish the observed relative frequency \(0.18\) from the probability in the underlying model.

Hints

- Identify what makes the household sample representative in the probability model. - Distinguish an observed sample proportion from the probability it estimates. - Recall what the law of large numbers says about repeated random trials.

Solution

1. The experiment is to select one household at random from the metropolitan-area population. The sampling method should give every household the same chance of selection. Under that model, the observed proportion \(0.18\) can be used to estimate the probability that a selected household has no car. 2. The relative frequency \(0.18\) is a value observed in one sample. The model probability is the long-run probability for the population process. Across many independent random selections under the same conditions, relative frequencies tend to stabilize near that probability, but one study’s value need not equal it exactly.

Answer

a) Select a household randomly so that every household has the same chance of selection; then \(0.18\) is an estimate of the model probability. b) The relative frequency is an observed sample value, while the model probability is the long-run value that relative frequencies tend to approach.

Experimental probability and simulations 41

A container holds an unknown number of red, blue, and green balls. In \(500\) draws with replacement, a red ball was drawn \(124\) times. a) Find the relative frequency of drawing a red ball and explain why it can be used to estimate the probability \(p\). b) Linh says, “There are three colors, so each color must have probability \(\frac{1}{3}\).” Evaluate this claim using the experimental result.

Hints

- Relate the number of red outcomes to the total number of trials. - Ask what information would be needed to make the three colors equally likely. - Compare the experimental estimate with the proposed theoretical value.

Solution

1. The relative frequency is \(\frac{124}{500}=0.248\). With many repeated draws under the same conditions, relative frequency tends to stabilize near the underlying probability, so \(0.248\) is a reasonable estimate of \(p\). 2. The value \(\frac{1}{3}\) would require the three colors to be equally likely, such as having equal numbers of balls of each color. That information is not given. The observed value \(0.248\) also differs substantially from \(\frac{1}{3}\), so the equal-probability claim is not justified.

Answer

a) \(0.248\); it is an experimental estimate of \(p\). b) The claim is not justified because three possible colors do not automatically make the colors equally likely, and the observed relative frequency is well below \(\frac{1}{3}\).

Experimental probability and simulations 42

A fair eight-sided die is labeled \(1\) through \(8\). 1. Let \(A\) be the event that the result is prime or a perfect square. List the outcomes in \(A\) and find \(P(A)\). 2. In \(250\) rolls, a prime occurred \(120\) times and a perfect square occurred \(65\) times. No result is both prime and a perfect square, so event \(A\) occurred \(185\) times. Find the relative frequency of \(A\) and compare it with the theoretical probability. 3. Explain how the relative frequency is expected to relate to the theoretical probability as the number of rolls increases.

Hints

- List the primes and perfect squares from \(1\) through \(8\). - Divide the number of occurrences of \(A\) by \(250\). - Recall what happens to relative frequency over many trials.

Solution

1. The primes are \(2,3,5,7\), and the perfect squares are \(1,4\). Thus, \(A=\{1,2,3,4,5,7\}\), so \(P(A)=\frac{6}{8}=0.75\). 2. The relative frequency is \(\frac{185}{250}=0.74\), which is \(0.01\) below the theoretical probability. 3. By the law of large numbers, the relative frequency tends to become more stable near the theoretical probability as the number of rolls increases.

Answer

1. \(A=\{1,2,3,4,5,7\}\) and \(P(A)=0.75\) 2. The relative frequency is \(0.74\), which is close to \(0.75\). 3. With more rolls, the relative frequency tends to approach the theoretical probability.

Experimental probability and simulations 43

A spinner has \(10\) equal sections labeled \(1\) through \(10\). Let \(E\) be the event that the result is prime or odd. a) Find the theoretical probability \(P(E)\). b) In \(200\) spins, event \(E\) occurred \(132\) times. Find the relative frequency and its difference from the theoretical probability. c) A classmate says, “Because the relative frequency is not exactly the theoretical probability, the spinner must be unfair.” Is that conclusion justified by this experiment alone? Explain.

Hints

- Form the union of the prime and odd outcomes without double-counting. - Compare the observed proportion numerically with the theoretical probability. - Ask whether a fair random process must produce its theoretical proportion exactly in every finite set of trials.

Solution

1. The primes are \(2,3,5,7\), and the odd numbers are \(1,3,5,7,9\). Their union is \(\{1,2,3,5,7,9\}\), so \(P(E)=\frac{6}{10}=0.6\). 2. The relative frequency is \(\frac{132}{200}=0.66\), which is \(0.66-0.60=0.06\) above the theoretical probability. 3. A finite random experiment is not expected to match its theoretical probability exactly every time. 4. Therefore, the fact that \(0.66\neq0.60\) does not by itself prove the spinner is unfair. The experiment establishes the observed difference, not a deterministic fairness conclusion.

Answer

a) \(0.6\) b) \(0.66\), which is \(0.06\) above the theoretical probability. c) No. Finite experimental results can differ from theoretical probabilities because of random variation; this one mismatch does not prove unfairness.

Experimental probability and simulations 44

A six-sided die is rolled \(1200\) times, and a \(6\) occurs \(246\) times. a) Find the relative frequency of rolling a \(6\) and compare it with the theoretical probability for a fair die. b) If the die were fair, about how many sixes would you expect in \(1200\) rolls? Does the result give you reason to question whether the die is fair? Explain why it is not proof. c) Using the observed relative frequency as your estimate, about how many \(6\)s would you predict in another \(200\) rolls?

Hints

- How could you compare the observed number of sixes with what a fair die predicts? - What can repeated random trials tell you, and what can they not guarantee? - To estimate a future count, consider how a rate relates to the number of rolls.

Solution

1. The relative frequency is \(\frac{246}{1200}=0.205\). A fair die has \(P(6)=\frac{1}{6}\), which is smaller. 2. For a fair die, the expected count is \(1200\cdot\frac{1}{6}=200\) sixes. The observed \(246\) is \(46\) more than expected, a noticeable difference over many rolls. This gives reason to suspect the die may be unfair, but chance can still produce differences, so it is not proof. 3. Using the observed rate to predict future results gives \(200\cdot0.205=41\) sixes. This is an estimate, not a guarantee.

Answer

a) \(0.205\), compared with \(\frac{1}{6}\) for a fair die. b) About \(200\) sixes expected if fair; \(246\) observed is \(46\) more. This raises doubt about fairness but does not prove the die is unfair. c) About \(41\) sixes.

Experimental probability and simulations 45

Use random digits \(0\) through \(9\) to simulate two independent trials with success probability \(0.3\). Give one possible assignment of digits to “success” and “failure,” and find the probability of at least one success.

Hints

- Assign exactly \(3\) of the \(10\) equally likely digits to success. - Use the complement of two failures.

Solution

1. One possible assignment is success for \(\{0, 1, 2\}\) and failure for \(\{3, 4, 5, 6, 7, 8, 9\}\). 2. The probability of no success in two trials is \(0.7^2 = 0.49\). 3. Therefore, \(P(\text{at least one success}) = 1 - 0.49 = 0.51\).

Answer

For example, assign \(0\), \(1\), and \(2\) to success; \(P(\text{at least one success}) = 0.51\).

Experimental probability and simulations 46

A warehouse robot sometimes turns clockwise after meeting an obstacle. Two tests record its turns. Test A: \(18\) clockwise turns in \(30\) trials. Test B: \(174\) clockwise turns in \(300\) trials. Find each experimental probability. Which test should usually provide the more stable estimate of the robot's long-run probability?

Hints

- Convert each result into a relative frequency. - Compare the amount of evidence behind each estimate. - Think about how random variation behaves over many trials.

Solution

1. Test A gives \(\frac{18}{30}=0.60\). 2. Test B gives \(\frac{174}{300}=0.58\). 3. Test B uses ten times as many trials. 4. Larger numbers of trials usually reduce random fluctuation, so Test B should provide the more stable estimate.

Answer

Test A: \(0.60\). Test B: \(0.58\). Test B should usually provide the more stable estimate.

Experimental probability and simulations 47

Use random digits to simulate whether at least one of three buses is late. For each bus, digits \(0\) and \(1\) mean “late,” and digits \(2\) through \(9\) mean “on time.” One trial uses three digits. In \(200\) simulated trials, at least one bus was late in \(99\) trials. Estimate the probability that at least one bus is late. Describe which three-digit outcomes count as successes.

Hints

- Connect each digit category with the event it represents. - Decide what must happen somewhere in a three-digit trial for the compound event to occur. - Use the successful-trial count as a relative frequency.

Solution

1. A trial is a success when at least one of its three digits is \(0\) or \(1\). 2. The experimental probability is \(\frac{99}{200}=0.495\). 3. Therefore, the simulation estimates the probability at about \(49.5\%\).

Answer

Any three-digit outcome containing at least one \(0\) or \(1\) counts as a success. The estimated probability is \(0.495\), or \(49.5\%\).

Experimental probability and simulations 48

An app is supposed to select one of four study prompts with equal probability. In \(80\) selections, the prompts appeared this many times: Prompt A: \(14\) Prompt B: \(18\) Prompt C: \(20\) Prompt D: \(28\) Find the experimental probability of Prompt D and compare it with the theoretical probability. Give one reasonable explanation for the difference.

Hints

- Use the observed count and total trials for the experimental value. - Determine the model probability from the number of equally likely choices. - Consider both chance variation and a possible problem with the model.

Solution

1. The experimental probability of Prompt D is \(\frac{28}{80}=0.35\). 2. With four equally likely prompts, the theoretical probability is \(\frac{1}{4}=0.25\). 3. The experimental probability is \(0.35-0.25=0.10\) greater. 4. The difference could be caused by random variation, or the app may not actually select all prompts equally.

Answer

The experimental probability is \(0.35\), compared with the theoretical probability \(0.25\). The difference is \(0.10\). Random variation or an unequal-selection bias in the app could explain it.

Experimental probability and simulations 49

A robot arm attempts to stack a block. Its cumulative results are shown. <table> <tr><th>Total attempts</th><th>Successful stacks</th></tr> <tr><td>\(10\)</td><td>\(8\)</td></tr> <tr><td>\(50\)</td><td>\(31\)</td></tr> <tr><td>\(200\)</td><td>\(119\)</td></tr> </table> Find the experimental probability at each stage. Which value is the best estimate of the long-run success probability, and why?

Hints

- Calculate a relative frequency for each cumulative result. - Compare how much evidence supports each value. - Consider which result should be least sensitive to a few unusual outcomes.

Solution

1. After \(10\) attempts, the experimental probability is \(\frac{8}{10}=0.80\). 2. After \(50\) attempts, it is \(\frac{31}{50}=0.62\). 3. After \(200\) attempts, it is \(\frac{119}{200}=0.595\). 4. The value \(0.595\) is the best estimate because it is based on the largest number of trials and should be less affected by short-run variation.

Answer

The experimental probabilities are \(0.80\), \(0.62\), and \(0.595\). The best long-run estimate is \(0.595\).

Experimental probability and simulations 50

A motion sensor gave a false alert \(84\) times. Its experimental probability of a false alert was \(0.35\). How many total trials were conducted?

Hints

- Relate the observed event count to the total number of trials. - Treat the given probability as a ratio. - Check that the recovered total makes the relative frequency correct.

Solution

1. Let \(n\) be the total number of trials. 2. The experimental probability gives \(\frac{84}{n}=0.35\). 3. Solving, \(n=84\div0.35=240\).

Answer

There were \(240\) trials.

Experimental probability and simulations 51

A simulation uses two-digit random numbers from \(00\) through \(99\) to model a launch cancellation probability of \(45\%\). Give a valid assignment of random numbers to “canceled” and “not canceled.” In \(120\) simulated launches, \(58\) were canceled. What experimental probability did the simulation produce?

Hints

- Match the desired percent to a count out of \(100\) equally likely numbers. - Keep the assigned ranges nonoverlapping and complete. - Use the simulated event count as a relative frequency.

Solution

1. One valid assignment is \(00\) through \(44\) for “canceled” and \(45\) through \(99\) for “not canceled.” 2. This assigns \(45\) of the \(100\) equally likely numbers to cancellation. 3. The experimental probability is \(\frac{58}{120}=\frac{29}{60}\).

Answer

One valid assignment is \(00\)–\(44\) for “canceled” and \(45\)–\(99\) for “not canceled.” The experimental probability is \(\frac{29}{60}\).

Experimental probability and simulations 52

Mateo wants to simulate an event with probability \(\frac{3}{8}\). Mateo generates one random digit from \(0\) through \(9\) and calls \(0\), \(1\), or \(2\) a success. Explain why this simulation is not correct. Give a correct simulation using a fair eight-sided die labeled \(1\) through \(8\).

Hints

- Compare the number of equally likely outcomes in the proposed device with the target probability. - Check what fraction of the proposed outcomes are labeled as successes. - Look for a random device whose equally likely outcomes make the target denominator natural.

Solution

1. The random-digit plan assigns \(3\) of \(10\) equally likely digits to success, so it models probability \(\frac{3}{10}\), not \(\frac{3}{8}\). 2. With a fair eight-sided die, designate outcomes \(1,\ 2,\ 3\) as success and \(4,\ 5,\ 6,\ 7,\ 8\) as failure. 3. This gives \(3\) successful outcomes out of \(8\), matching \(\frac{3}{8}\).

Answer

The digit plan models \(\frac{3}{10}\), so it is incorrect. A correct plan is to roll a fair eight-sided die and count \(1,\ 2,\) or \(3\) as success.

Experimental probability and simulations 53

Two independent computer simulations estimate the same event. Simulation A: \(86\) successes in \(200\) trials Simulation B: \(126\) successes in \(300\) trials Find each estimate and the combined estimate using all \(500\) trials.

Hints

- Calculate each relative frequency separately. - For a combined estimate, combine both the successful outcomes and the trials. - Check that the combined value lies between the two individual estimates.

Solution

1. Simulation A gives \(\frac{86}{200}=0.43\). 2. Simulation B gives \(\frac{126}{300}=0.42\). 3. Together there are \(86+126=212\) successes in \(200+300=500\) trials. 4. The combined estimate is \(\frac{212}{500}=0.424\).

Answer

Simulation A: \(0.43\). Simulation B: \(0.42\). Combined estimate: \(0.424\).

Experimental probability and simulations 54

A game token lands on its star side \(22\) times in \(40\) trials. In a new set of \(4\) trials, it lands on the star side all \(4\) times. Freyja uses only the new set and claims the probability is \(1\). Find the experimental probability using all the trials, and explain why it is more reliable.

Hints

- Combine the evidence from both sets rather than discarding the larger set. - Compare how much data supports the two estimates. - Consider how a short streak can distort a small-sample result.

Solution

1. Across both sets, the token lands on the star side \(22+4=26\) times. 2. The total number of trials is \(40+4=44\). 3. The combined experimental probability is \(\frac{26}{44}=\frac{13}{22}\). 4. The combined estimate is more reliable because it uses many more trials and is less affected by the unusual run of four stars.

Answer

The combined experimental probability is \(\frac{13}{22}\). It is more reliable because it is based on \(44\) trials instead of \(4\).

Experimental probability and simulations 55

A digital music wheel has three outcomes: jazz, rock, and pop. The wheel's probabilities are unknown. In \(180\) trials, jazz occurred \(72\) times, rock \(54\) times, and pop \(54\) times. Estimate the probability of each outcome. Based on the experiment, predict the numbers of each outcome in the next \(50\) trials.

Hints

- Divide each outcome count by the same total number of trials. - Check that the estimated probabilities add to \(1\). - Apply each estimated rate to the new trial count.

Solution

1. The estimated jazz probability is \(\frac{72}{180}=0.40\). 2. The estimated rock probability is \(\frac{54}{180}=0.30\). 3. The estimated pop probability is \(\frac{54}{180}=0.30\). 4. In \(50\) trials, the predictions are \(0.40\cdot50=20\) jazz, \(0.30\cdot50=15\) rock, and \(0.30\cdot50=15\) pop.

Answer

Estimated probabilities: jazz \(0.40\), rock \(0.30\), pop \(0.30\). Predicted counts in \(50\) trials: \(20\) jazz, \(15\) rock, and \(15\) pop.

Experimental probability and simulations 56

A probability model says an event should occur with probability \(0.40\). Simulation A records \(18\) occurrences in \(50\) trials. Simulation B records \(390\) occurrences in \(1000\) trials. Find each experimental probability and its absolute difference from the model probability. Which simulation agrees more closely with the model?

Hints

- Convert each frequency into a probability estimate. - Measure how far each estimate is from the model value. - Compare the sizes of those differences.

Solution

1. Simulation A gives \(\frac{18}{50}=0.36\), which differs from \(0.40\) by \(0.04\). 2. Simulation B gives \(\frac{390}{1000}=0.39\), which differs from \(0.40\) by \(0.01\). 3. Simulation B agrees more closely because \(0.01<0.04\).

Answer

Simulation A: probability \(0.36\), difference \(0.04\). Simulation B: probability \(0.39\), difference \(0.01\). Simulation B agrees more closely with the model.

Experimental probability and simulations 57

A probability model gives an event probability of \(0.65\). In \(400\) trials, the event occurred \(244\) times. Find the expected count from the model, the experimental probability, and the difference between the theoretical and experimental probabilities.

Hints

- Apply the model rate to the number of trials. - Use the observed frequency to form an experimental rate. - Compare both the probabilities and the corresponding counts.

Solution

1. The model predicts \(0.65\cdot400=260\) occurrences. 2. The experimental probability is \(\frac{244}{400}=0.61\). 3. The probability difference is \(0.65-0.61=0.04\). 4. The observed count is \(260-244=16\) below the model prediction.

Answer

Expected count: \(260\). Experimental probability: \(0.61\). Probability difference: \(0.04\). The observed count is \(16\) below the model prediction.

Experimental probability and simulations 58

A simulation models three independent seeds, each with a \(70\%\) chance of germinating. A trial is a success when all three seeds germinate. In \(500\) trials, all three seeds germinated \(174\) times. Find the experimental probability and compare it with the theoretical probability.

Hints

- Use successful trials over total trials for the simulation result. - For the model, account for all three independent successes. - Compare the two probabilities by finding their distance apart.

Solution

1. The experimental probability is \(\frac{174}{500}=0.348\). 2. Because the three germination outcomes are independent, the theoretical probability is \(0.7\cdot0.7\cdot0.7=0.343\). 3. The difference is \(0.348-0.343=0.005\). 4. The simulation result is very close to the theoretical probability.

Answer

The experimental probability is \(0.348\), and the theoretical probability is \(0.343\). They differ by \(0.005\).

Experimental probability and simulations 59

A simulation produced \(45\) successes in \(120\) trials. Deniz reports the experimental probability as \(0.45\). Correct Deniz's calculation. Using the corrected probability, predict the number of successes in \(80\) more trials.

Hints

- Relate the number of successes to the total number of trials. - Check whether the reported decimal represents that relative frequency. - Use the corrected experimental rate to make the new prediction.

Solution

1. The experimental probability is \(\frac{45}{120}=\frac{3}{8}=0.375\), not \(0.45\). 2. The predicted number of successes is \(0.375\cdot80=30\).

Answer

The corrected experimental probability is \(0.375\). About \(30\) successes are predicted in \(80\) more trials.

Experimental probability and simulations 60

Five batches of \(100\) simulated trials produced these success counts: \(42,\ 48,\ 51,\ 46,\ 53\) Find the mean success count and the range of the batch counts. Use the mean to estimate the event probability, and explain what the range shows.

Hints

- Summarize the repeated batch results with a center and a spread. - Convert the typical count into a relative frequency. - Interpret the spread as trial-to-trial variation, not a change in the event itself.

Solution

1. The mean success count is \(\frac{42+48+51+46+53}{5}=\frac{240}{5}=48\). 2. The estimated probability is \(\frac{48}{100}=0.48\). 3. The range is \(53-42=11\) successes. 4. The range shows that separate batches can vary even when they simulate the same event.

Answer

The mean success count is \(48\), so the estimated probability is \(0.48\). The range is \(11\), showing noticeable random variation among batches.

Experimental probability and simulations 61

A bag contains seven cards of the same size, numbered \(1\) through \(7\), with each number appearing once. A class mixes the cards, draws one without looking, replaces it, and mixes again before the next draw. After \(350\) draws, card \(3\) was chosen \(55\) times. Find the experimental probability for card \(3\), compare it with the theoretical probability, and give the exact difference.

Hints

- Use the observed count over the total number of selections. - Determine the model probability from the number of equally likely choices. - Express both fractions with the same denominator to find their difference.

Solution

1. The experimental probability is \(\frac{55}{350}=\frac{11}{70}\). 2. The theoretical probability is \(\frac{1}{7}=\frac{10}{70}\). 3. The difference is \(\frac{11}{70}-\frac{10}{70}=\frac{1}{70}\). 4. The observed frequency is close to the equal-choice model, though more trials would provide stronger evidence.

Answer

Experimental probability: \(\frac{11}{70}\). Theoretical probability: \(\frac{1}{7}\). Difference: \(\frac{1}{70}\).

Experimental probability and simulations 62

A computer simulates four independent fair coin flips. In \(500\) trials, exactly one head occurred \(130\) times. Find the experimental probability. Then find the theoretical probability and compare the results.

Hints

- Use the simulated success count as a relative frequency. - Count how many equally likely sequences contain the required number of heads. - Compare the observed and model probabilities.

Solution

1. The experimental probability is \(\frac{130}{500}=0.26\). 2. There are \(2^4=16\) equally likely four-flip outcomes. 3. Exactly one head can occur in \(4\) positions, so the theoretical probability is \(\frac{4}{16}=0.25\). 4. The simulation differs from the model by \(0.26-0.25=0.01\).

Answer

The experimental probability is \(0.26\), and the theoretical probability is \(0.25\). They differ by \(0.01\).

Experimental probability and simulations 63

A model gives a rare event probability of \(0.02\). In \(60\) simulated trials, the event never occurred. Maarja concludes that the event is impossible. Evaluate the conclusion using the model's expected number of occurrences.

Hints

- Connect the model probability with the number of simulated trials. - Distinguish an average prediction from a guaranteed result. - Consider how much short-run variation a rare event can show.

Solution

1. The model predicts \(0.02\cdot60=1.2\) occurrences on average. 2. An expected count of \(1.2\) does not mean every set of \(60\) trials must contain an occurrence. 3. Getting \(0\) occurrences in a small set of trials can happen for a rare event. 4. Therefore, the simulation does not show that the event is impossible.

Answer

The conclusion is incorrect. The expected count is \(1.2\), but a rare event can still occur \(0\) times in \(60\) trials.

Experimental probability and simulations 64

A simulation uses random digits to imitate a fair six-sided number cube. Digits \(1\) through \(6\) represent the six outcomes. Digits \(0\), \(7\), \(8\), and \(9\) are ignored, and another digit is generated. Among \(240\) accepted trials, outcome \(6\) appeared \(46\) times. Find the experimental probability, the theoretical probability, and their exact difference.

Hints

- Focus only on trials that the simulation accepts. - Divide the accepted occurrences of \(6\) by all accepted trials. - Compare with one outcome among six equally likely values.

Solution

1. Ignoring the other digits leaves \(1\) through \(6\) as equally likely accepted outcomes. 2. The experimental probability of outcome \(6\) is \(\frac{46}{240}=\frac{23}{120}\). 3. The theoretical probability is \(\frac{1}{6}=\frac{20}{120}\). 4. The difference is \(\frac{23}{120}-\frac{20}{120}=\frac{3}{120}=\frac{1}{40}\).

Answer

Experimental probability: \(\frac{23}{120}\). Theoretical probability: \(\frac{1}{6}\). Difference: \(\frac{1}{40}\).

Experimental probability and simulations 65

A simulation is run in two equal parts. First \(200\) trials: \(90\) successes Next \(200\) trials: \(110\) successes Find the experimental probability for each part and for all \(400\) trials. What do the results show about random variation?

Hints

- Compute a relative frequency for each section of the data. - Combine the event counts and trial counts for the overall result. - Compare the smaller-run estimates with the larger combined estimate.

Solution

1. The first part gives \(\frac{90}{200}=0.45\). 2. The second part gives \(\frac{110}{200}=0.55\). 3. Together there are \(90+110=200\) successes in \(400\) trials. 4. The overall experimental probability is \(\frac{200}{400}=0.50\). 5. Equal-sized parts can differ noticeably even when the combined long-run estimate is centered between them.

Answer

First part: \(0.45\). Second part: \(0.55\). Overall: \(0.50\). The two parts show short-run variation around the combined estimate.

Experimental probability and simulations 66

A simulation must model an event with probability \(12.5\%\). It generates a random three-digit number from \(000\) through \(999\). Give a valid range of numbers for “success.” In \(400\) trials, the simulation produced \(46\) successes. Find the experimental probability.

Hints

- Convert the target percent into a count out of \(1000\). - Choose a consecutive block containing exactly that many outcomes. - Use the simulated success count as a relative frequency.

Solution

1. Since \(12.5\%=0.125=\frac{125}{1000}\), one valid assignment is \(000\) through \(124\) for success. 2. That range contains \(125\) of the \(1000\) equally likely numbers. 3. The experimental probability is \(\frac{46}{400}=0.115\).

Answer

One valid assignment is \(000\)–\(124\) for success. The experimental probability is \(0.115\).

Experimental probability and simulations 67

An event occurred \(322\) times in \(800\) simulated trials. Which theoretical model is most consistent with the result: \(\frac{1}{3}\), \(\frac{2}{5}\), or \(\frac{1}{2}\)? Show the comparisons.

Hints

- Convert the observed frequency into a decimal probability. - Express each candidate model in a comparable form. - Select the model with the smallest absolute difference.

Solution

1. The experimental probability is \(\frac{322}{800}=0.4025\). 2. Its distance from \(\frac{1}{3}\approx0.3333\) is about \(0.0692\). 3. Its distance from \(\frac{2}{5}=0.4\) is \(0.0025\). 4. Its distance from \(\frac{1}{2}=0.5\) is \(0.0975\). 5. The model \(\frac{2}{5}\) is closest to the experimental result.

Answer

The model \(\frac{2}{5}\) is most consistent with the simulation because \(0.4025\) is only \(0.0025\) away from \(0.4\).

Experimental probability and simulations 68

After \(400\) simulated trials, an event has experimental probability \(0.275\). How many successes have occurred? Then \(50\) more trials produce no successes. What is the updated experimental probability? Give the new probability as a fraction in simplest form.

Hints

- Convert the initial relative frequency back into a count. - Update the total trial count without changing the success count. - Recalculate the relative frequency using the new totals.

Solution

1. The number of successes is \(0.275\cdot400=110\). 2. After \(50\) more trials, there are still \(110\) successes in \(450\) trials. 3. The updated experimental probability is \(\frac{110}{450}=\frac{11}{45}\).

Answer

There were \(110\) successes after \(400\) trials. The updated experimental probability is \(\frac{11}{45}\).

Experimental probability and simulations 69

Eight classes each run \(50\) trials of the same simulation. Their success counts are: \(28,\ 31,\ 27,\ 30,\ 29,\ 32,\ 25,\ 28\) Find the combined experimental probability. Also find the mean success count per class.

Hints

- Combine all event counts before forming the overall relative frequency. - Count the total number of trials across the equal-sized groups. - Use the same total success count to find the per-group average.

Solution

1. The total number of successes is \(28+31+27+30+29+32+25+28=230\). 2. The total number of trials is \(8\cdot50=400\). 3. The combined experimental probability is \(\frac{230}{400}=0.575\). 4. The mean success count per class is \(230\div8=28.75\).

Answer

The combined experimental probability is \(0.575\), and the mean success count is \(28.75\) per class.

Experimental probability and simulations 70

Ibrahima says, “After exactly \(50\) trials, the experimental probability was exactly \(0.33\).” Explain why this cannot be exact. Give the two closest possible experimental probabilities.

Hints

- Think about what the numerator of an experimental probability represents. - Ask whether the reported decimal corresponds to a whole-number count of successes. - Check the neighboring possible success counts.

Solution

1. An exact experimental probability after \(50\) trials must have the form \(\frac{k}{50}\), where \(k\) is a whole-number success count. 2. An exact probability of \(0.33\) would require \(0.33\cdot50=16.5\) successes, which is impossible. 3. The nearest whole-number counts are \(16\) and \(17\). 4. The corresponding probabilities are \(\frac{16}{50}=0.32\) and \(\frac{17}{50}=0.34\).

Answer

It cannot be exact because it would require \(16.5\) successes. The two closest possible probabilities are \(0.32\) and \(0.34\).

Experimental probability and simulations 71

An experiment estimates an event probability as \(0.18\). About how many trials are needed to expect \(90\) occurrences? If \(104\) occurrences actually happen in that many trials, what is the new experimental probability?

Hints

- Relate the expected event count to the estimated rate and trial total. - Solve for the missing trial count. - Use the actual event count to update the relative frequency.

Solution

1. Let \(n\) be the number of trials. The estimate gives \(0.18n=90\). 2. Solving, \(n=90\div0.18=500\). 3. With \(104\) occurrences in \(500\) trials, the new experimental probability is \(\frac{104}{500}=0.208\).

Answer

About \(500\) trials are needed. The new experimental probability is \(0.208\).

Experimental probability and simulations 72

A factory checks parts made by three machines. Each machine is expected to have a defect rate of \(5\%\). The inspection gives these results: - Machine A: \(12\) defective parts out of \(200\) inspected. - Machine B: \(105\) defective parts out of \(2000\) inspected. - Machine C: \(1160\) defective parts out of \(20{,}000\) inspected. a) Find the relative frequency of defective parts for each machine. b) Based on these data, which machine gives the strongest indication that its defect probability differs from \(0.05\)? Explain using the law of large numbers.

Hints

- Divide the number of defective parts by the number inspected. - Compare each relative frequency with \(0.05\). - A deviation based on many trials is generally less likely to be caused only by random variation.

Solution

1. a) Machine A: \(\frac{12}{200} = 0.06\). Machine B: \(\frac{105}{2000} = 0.0525\). Machine C: \(\frac{1160}{20000} = 0.058\). 2. The absolute differences from \(0.05\) are \(0.01\), \(0.0025\), and \(0.008\), respectively. 3. Random variation usually has less effect when many parts are inspected. Machine C still has a relative frequency noticeably above \(0.05\) after \(20{,}000\) inspections, so its data give the strongest indication of a different defect probability. This is evidence, not proof.

Answer

a) Machine A: \(0.06\); Machine B: \(0.0525\); Machine C: \(0.058\) b) Machine C, because its relative frequency remains noticeably above \(0.05\) despite the very large number of inspections.

Experimental probability and simulations 73

Two different simulations model an event with theoretical probability \(\frac{3}{4}\). Simulation A uses a fair four-section spinner with three success sections and gives \(152\) successes in \(200\) trials. Simulation B flips two independent fair coins and counts every outcome except two tails as a success. It gives \(147\) successes in \(200\) trials. Find each experimental probability and the combined estimate.

Hints

- Verify that each method assigns the intended fraction of outcomes to success. - Calculate each observed relative frequency. - Combine counts rather than averaging without considering trial totals.

Solution

1. Simulation A gives \(\frac{152}{200}=0.76\). 2. Simulation B is valid because \(3\) of the \(4\) equally likely two-coin outcomes are successes, and its experimental probability is \(\frac{147}{200}=0.735\). 3. Together there are \(152+147=299\) successes in \(400\) trials. 4. The combined estimate is \(\frac{299}{400}=0.7475\), close to \(\frac{3}{4}=0.75\).

Answer

Simulation A: \(0.76\). Simulation B: \(0.735\). Combined estimate: \(0.7475\).

Experimental probability and simulations 74

A computer simulates three independent shots, each with a hit probability of \(0.60\). A trial is a success when at least two shots hit. In \(1000\) trials, there were \(645\) successes. Find the experimental probability and compare it with the theoretical probability.

Hints

- Separate the successful compound event into its possible cases. - Find the model probability for each case before combining them. - Compare the simulation result with the total model probability.

Solution

1. The experimental probability is \(\frac{645}{1000}=0.645\). 2. Exactly two hits has probability \(3\cdot(0.6)^2\cdot0.4=0.432\). 3. Three hits has probability \((0.6)^3=0.216\). 4. The theoretical probability is \(0.432+0.216=0.648\). 5. The experimental and theoretical probabilities differ by \(0.648-0.645=0.003\).

Answer

The experimental probability is \(0.645\), and the theoretical probability is \(0.648\). They differ by \(0.003\).

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Grade 7 math worksheets

Experimental probability and simulations

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