7th Grade Math Quiz: Design Simulations For Compound Events

What this quiz covers

This quiz focuses on Design Simulations For Compound Events, giving you a quick way to practice the rules, question types, and explanations that matter most for 7th Grade Math.

How to use this quiz

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

Question 1

1. Roll a die once per trial and count a win if the roll is even; repeat 100 trials.
2. Roll a die twice per trial and count a win if both rolls are even; repeat 100 trials and estimate $P(\text{win})$ as wins/100. (correct answer)
3. Roll a die twice total, and if you get at least one even, estimate the probability as $1/2$.
4. Flip a coin twice per trial and count a win if both are heads; repeat 100 trials.

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency, for winning by rolling two even numbers on a die (P(even)=0.5). Design: (1) identify the event as two independent rolls both even, (2) use a die (even on 2,4,6 for P=3/6=0.5), (3) define success as both even in two rolls per trial, (4) perform many trials (e.g., 100), (5) estimate P as wins over trials. Example: roll a die twice per trial, check if both are even, repeat 100 times, and calculate the proportion of wins to estimate the 0.25 probability. Choice B is correct, using the die to match the game, compounding two rolls per trial, defining the both-even success, and running 100 trials. Errors include single rolls, mismatched devices or probabilities, or insufficient trials like only two total rolls. In designing, examine the event structure, choose a matching random device, map outcomes properly, specify trial numbers, and outline frequency estimation. Run simulations by randomizing actions, recording results, counting successes, and computing estimates; avoid mismatches, low trials, incorrect mappings, or errors in calculation.

Question 2

1. About 25 students (correct answer)
2. About 4 students
3. About 40 students
4. About 10 students

Explanation: The chance of choosing pizza is 40%, or 0.4, so on average you would expect 1 pizza choice for every 2.5 students checked, since 1 divided by 0.4 equals 2.5. To reach 10 pizza choices, multiply 10 by 2.5, or equivalently divide 10 by 0.4, which gives 25 students. This matches Choice A. Choice B, about 4 students, comes from multiplying 10 by 0.4 instead of dividing, which answers a different question. Choice C, about 40 students, and Choice D, about 10 students, do not match the correct expected-value calculation for reaching 10 successes at a 40% success rate.

Question 3

1. $1.22$
2. $0.40$
3. $0.82$ (correct answer)
4. $0.18$

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency. A proper design involves: (1) identifying the event probability, here 0.4 for a correct guess, (2) choosing a matching device, (3) defining success as at least 1 correct in 4 attempts per trial, (4) running trials like 100, and (5) estimating P as successes over total trials, such as 82/100=0.82. For example, in simulating P=0.4 success, run 100 trials of 4 attempts each, count trials with at least 1 success, and if 82, estimate P=0.82. The correct estimate is choice C, which is 82/100=0.82, directly from the simulation results. Errors in other choices include incorrect calculations like 0.18 (perhaps 1-0.82) in A, the base P=0.4 in B, and invalid 1.22 in D. When designing such simulations, first analyze the compound event and its base probability, then select a device that matches it and map outcomes to define success. Finally, run sufficient trials by randomizing, record and count successes, calculate the estimate as frequency, and avoid mistakes like arithmetic errors in estimation.

Question 4

1. Use a 10-section spinner with 4 sections labeled A and 6 labeled not A. For each trial, spin 10 times (one per donor), record how many A's occur, and repeat for 100 trials. (correct answer)
2. Pick 10 students you know and count how many have type A. Repeat until you feel confident.
3. Flip a coin once for each donor (heads = type A). Do 10 flips total and use the result as the estimate.
4. Roll a die once for each donor (1-4 = type A). Do 10 rolls total and use the result as the estimate.

Explanation: Since about 40 percent of students have blood type A, the simulation device needs to match that probability, and a 10 section spinner with 4 sections labeled A does exactly that, making choice A correct when spun 10 times per trial and repeated for 100 trials. Choice B relies on picking students the nurse already knows, which is not a random method at all and would likely be biased. Choice C uses a coin, which represents a probability of 0.5, not 0.4, so it does not match the given probability. Choice D uses a die with 4 of 6 sides marking type A, representing a probability of about 0.67, another mismatch. Repeating the simulation across many trials, such as 100, rather than relying on just one round of 10 actions, is also necessary to get a reliable estimate of the probability.

Question 5

1. $\frac{59}{120}\approx 0.49$
2. $\frac{61}{120}\approx 0.51$ (correct answer)
3. $\frac{120}{61}\approx 1.97$
4. $\frac{61}{240}\approx 0.25$

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency. The design involves: (1) identifying the event as making both free throws (each $P=0.7$ via $7/10$ spinner), (2) choosing the spinner, (3) defining success as two makes per trial, (4) running 120 trials, (5) estimating $P$ as successes/120 ($61/120\approx0.508$). For example, with 61 successes in 120 trials, the estimate is $61/120\approx0.51$; expected successes in 120 trials at theoretical $P=0.49$ is about 59, close to 61. The correct calculation is option B, which properly computes the frequency $61/120\approx0.51$. Errors in other options include inverting the fraction ($120/61$), using non-successes ($59/120$), or wrong numerator ($61/240$ halves it). When designing such simulations: (1) analyze the event, (2) select device, (3) map success, (4) run trials, (5) estimate by ratio. Running includes recording successes and calculating correctly; mistakes are arithmetic errors like wrong fractions or misinterpreting data.

Question 6

1. About 10 trials
2. About 4 trials
3. About 25 trials (correct answer)
4. About 40 trials

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency, specifically calculating expected trials for about 10 successes at P=0.4. The formula for expected trials to reach k successes in Bernoulli trials is approximately k/P, so for 10 chocolate choices at P=0.4, expect 10/0.4=25 trials. Example: if simulating daily choices with P=0.4, the teacher should anticipate around 25 days to observe about 10 chocolate selections based on the inverse of probability. The correct answer is C, about 25 trials, directly from the calculation 10/0.4=25. Common errors involve wrong applications like kP (4), 1/P (2.5 but not matching choices), or Pk (4), or misreading as total successes needed. When incorporating expected trials in designs, analyze success probability, use k/P for expectation, integrate into simulation planning for sufficient runs. Mistakes often include arithmetic errors or confusing expected value with other metrics; verify by recomputing k divided by P.

Question 7

1. Flip a coin once per turn (heads = event) and repeat 100 turns.
2. Use a 10-section spinner with 4 sections labeled Move and 6 labeled Not Move, then roll a die. Count a success only if the spinner lands on Move and the die shows 6. Repeat 100 turns. (correct answer)
3. Use only a die: roll once per turn; if it’s 1–2 count it as Move, and if it’s 6 count it as rolling a 6. Repeat 100 turns.
4. Use a spinner with 6 equal sections labeled 1–6. Spin once per turn; if it lands on 6, count it as both Move and roll a 6. Repeat 100 turns.

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency, for the AND event of Move card (P=0.4) and rolling 6 (P=1/6). Design: (1) identify independent events, (2) use separate devices like 10-spinner (4 Move for P=0.4) and die, (3) success if both occur per trial, (4) run 100 trials, (5) estimate as successes/trials. Example: spin for Move, roll die, success only if Move and 6, repeat 100 times for frequency approximating 0.4*(1/6)≈0.067. Choice B is correct, using spinner for 0.4, die for 1/6, combining for AND, and 100 trials. Errors include single devices without compounding, mismatched P (e.g., 1-2 on die=1/3), or incorrect success definitions. Design by breaking into parts, choosing devices per probability, mapping to joint success, specifying trials, outlining estimation. Execute with independent randomizations, result tracking, counting, calculating; pitfalls are mismatches, insufficient trials, logic errors in AND, arithmetic.

Question 8

1. Choose a number from 1 to 6 without looking. If it is a 1, count a success. Repeat 10 trials and estimate the probability.
2. Roll a die 4 times per trial. Count a success if at least one roll is a 1. Repeat 150 trials and estimate the probability as successes/150. (correct answer)
3. Roll a die 4 times total. If any roll is a 1, say the probability is 1.
4. Roll a die once per trial. Count a success if it is a 1. Repeat 150 trials and use that as the probability of getting at least one 1 in 4 rolls.

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency, for at least one 1 in 4 die rolls (P(1)=1/6). Design: (1) recognize binomial with P=1/6, (2) use die (1 as success), (3) define at-least-one in 4 rolls per trial, (4) run many trials (e.g., 150), (5) estimate P as successes/trials. Example: roll die 4 times per trial, check for any 1, repeat 150 times, compute proportion for estimate near 1-(5/6)^4≈0.518. Choice B is best, using die, compounding 4 rolls, specifying at-least-one, and 150 trials. Common errors are total rolls without trials, equating single roll to compound, or low repetitions. In designing, analyze structure, select device, map to compound outcome, set trial count, describe frequency method. Run by randomizing rolls, recording, counting successes, calculating ratios; avoid mismatches, few trials, mapping flaws, errors in computation.

Question 9

1. Roll a number cube once; let 1–4 mean type A.
2. Flip a coin once; let heads mean type A.
3. Pick a card from a standard 52-card deck; let hearts mean type A.
4. Use a 10-section spinner with 4 sections labeled A and 6 labeled not A; spin 100 times; estimate P(\text{A})=\frac{\text{# of A spins}}{100}. (correct answer)

Explanation: This question tests designing simulations using random devices to estimate compound event probabilities through trials and frequency. The design involves: (1) identifying the event as the probability that the first student checked has type A blood (P≈0.4), (2) choosing a device like a 10-section spinner with 4 sections labeled A (P=4/10=0.4), (3) defining success as landing on A, (4) running multiple trials such as 100 spins, (5) estimating P as the number of successes divided by trials. For example, with a 10-section spinner (4 A sections), after 100 spins with 42 A's, the estimate is 42/100=0.42; for expected trials to get 10 successes at P=0.4, it's about 10/0.4=25 trials. The correct design is option B, which uses a matching spinner, spins 100 times, and estimates properly for the single event of the first student having type A. Errors in other options include mismatched probabilities (coin is 0.5, cube 4/6≈0.67, cards 13/52=0.25) and insufficient trials or improper setup. When designing such simulations: (1) analyze the event and its probability, (2) select a device that matches the probability, (3) map outcomes to define success, (4) specify a sufficient number of trials, (5) describe how to estimate by frequency. Common mistakes include probability mismatches, too few trials, undefined mappings, or arithmetic errors in estimation.

Question 10

1. Flip a coin 5 times (heads = defective). Repeat 100 trials.
2. Roll a die 5 times (1 = defective). Do this once and use that single result as the probability.
3. Choose 5 pens from your pencil case and see if any seem defective. Use that as the estimate.
4. Use a 10-section spinner with 1 section labeled Defective and 9 labeled OK. For each trial, spin 5 times and record whether at least 1 Defective occurs. Repeat 100 trials and estimate with successes/trials. (correct answer)

Explanation: To simulate this compound event, you need a device where each trial matches the real probability of 10% defective. A 10-section spinner with 1 section for Defective and 9 for OK works because 1/10 equals 0.1, the same rate as the pens. Spinning it 5 times per trial models checking 5 pens, and repeating this 100 times gives enough data for a reliable estimate. Choice A uses a coin, which represents a 50% chance, not 10%, so it does not match the problem. Choice B uses a die with a 1 in 6 chance, and Choice C is not random at all since it relies on guessing which pens look defective.

Question 11

1. Use a die; 1-2 = Success and 3-6 = Fail. Do 5 trials and estimate P as successes/5.
2. Use a 10-section spinner with 6 sections labeled Success and 4 labeled Fail. Spin once and use that to decide the outcome for all 3 launches in the trial. Repeat 100 trials.
3. Flip a coin 3 times; heads = Success. Repeat 100 trials and estimate P as successes/100.
4. Use a 10-section spinner with 6 sections labeled Success and 4 labeled Fail. Spin 3 times per trial, count trials with exactly 2 Successes, repeat 100 trials, and estimate P as successes/100. (correct answer)

Explanation: To match the real probability of 60% success, use a device where 6 out of 10 outcomes represent Success, such as a 10-section spinner with 6 sections labeled Success and 4 labeled Fail. Each trial should model all 3 launches by spinning 3 times, then counting how many trials result in exactly 2 Successes out of those 3 spins. Repeating this for 100 trials gives a reliable estimate of the probability. Choice A uses a die with only 2 out of 6 sides marked Success, which is about 33%, not 60%, and it only uses 5 trials, which is too few for a solid estimate. Choice B spins only once and applies that single result to all 3 launches, which does not simulate 3 independent attempts. Choice C uses a coin, giving a 50% chance instead of 60%.

Question 12

1. Roll a die; let 1-2 = Make and 3-6 = Miss. Roll 4 times per trial and count trials with at least 3 Makes over 100 trials.
2. Use a 10-section spinner with 7 sections labeled Make and 3 labeled Miss. Spin 4 times per trial and count trials with 3 or 4 Makes over 100 trials. (correct answer)
3. Use a 10-section spinner with 3 sections labeled Make and 7 labeled Miss. Spin 4 times per trial and count trials with at least 3 Makes over 100 trials.
4. Flip a coin; let heads = Make. Flip 4 times per trial and count trials with at least 3 heads over 100 trials.

Explanation: To match a 70% chance of success, use a device where 7 out of 10 outcomes count as a Make, such as a 10-section spinner with 7 sections labeled Make and 3 labeled Miss. Each trial should model 4 free throws by spinning 4 times, then checking whether the trial has 3 or 4 Makes. Running 100 trials gives enough data for a reliable probability estimate. Choice A uses a die where only 2 out of 6 sides represent a Make, which is about 33%, not 70%. Choice C reverses the spinner so only 3 out of 10 sections are Make, giving about 30% instead of 70%. Choice D uses a coin, which gives a 50% chance instead of 70%.

Question 13

1. Use a 10-section spinner: 6 sections labeled G and 4 sections labeled N.
2. Use a 10-section spinner: 4 sections labeled G and 6 sections labeled N. (correct answer)
3. Use a die: 1-4 = graphic novel, 5-6 = not graphic novel.
4. Use a coin: heads = graphic novel, tails = not graphic novel.

Explanation: Since about 40% of library books are graphic novels, the simulation device needs 4 out of 10 outcomes to represent a graphic novel, such as a 10-section spinner with 4 sections labeled G and 6 labeled N. This matches the 40% chance exactly, since 4/10 = 0.4. Choice A reverses the spinner, using 6 sections for G and 4 for N, which represents 60% instead of 40%. Choice C uses a die with 4 out of 6 sides for graphic novel, which is about 67%, far too high. Choice D uses a coin, giving a 50% chance instead of 40%.

Question 14

1. Roll a die 5 times; let 1-4 = sore throat. Repeat 10 trials and estimate $P$ as successes/10.
2. Flip a coin 5 times; let heads = sore throat. Count trials with exactly 2 heads, and estimate the probability after 100 trials.
3. Use a 10-section spinner with 4 sections labeled S (sore throat) and 6 labeled N. Spin 5 times per trial, record whether exactly 2 S occur, repeat 100 trials, and estimate $P$ as (number of successful trials)/100. (correct answer)
4. Use a 10-section spinner with 6 sections labeled S and 4 labeled N. Spin 5 times and use the result of the first spin only to decide if the trial is a success.

Explanation: Since about 40 percent of students have a sore throat, the simulation device needs to match that probability, and a 10 section spinner with 4 sections labeled S does exactly that, making choice C correct. Choice A uses a die with 4 of 6 faces marking sore throat, which represents a probability of about 67 percent, not 40 percent. Choice B uses a coin, representing a 50 percent probability, another mismatch. Choice D flips the spinner sections, using 6 S and 4 N, and only uses the first spin, so it neither matches the probability nor tests all 5 students per trial. Only choice C correctly models both the 40 percent probability and the 5 student trial structure.

Question 15

1. Use a 10-section spinner with 4 bonus sections and a number cube; in each trial, spin once and roll once; success if bonus AND roll a 1; repeat 100 trials; estimate probability by successes/100. (correct answer)
2. Flip a coin twice per trial; success if both flips are heads.
3. Roll a number cube once per trial; success if you roll a 1 or 2 (to match 0.4).
4. Use a 10-section spinner with 4 bonus sections; spin once and if it lands on bonus, count it as success (no die roll needed).

Explanation: The situation has two independent stages, drawing a bonus card with probability 0.4 and rolling a 1 with probability about one sixth, so the simulation needs one device matched to each stage, which is exactly what choice A does with a 10 section spinner and a number cube. Choice B uses two coin flips, which does not represent either stage's probability. Choice C uses only the number cube and ignores the bonus card stage entirely, so it cannot model the compound event. Choice D uses only the spinner and ignores the number cube roll, also missing one of the two stages. Only choice A pairs a device with each stage and defines success as both events happening together in the same trial.

Question 16

1. $\frac{39}{6}=6.5$
2. $\frac{39}{120}=0.325$ (correct answer)
3. $\frac{39}{2}=19.5$
4. $\frac{120}{39}\approx 3.08$

Explanation: Since 39 of the 120 trials resulted in at least one 6, the estimated probability is the number of successful trials divided by the total number of trials, which is 39 divided by 120, equal to 0.325, matching choice B. Choice A, 39/6=6.5, incorrectly divides by 6 instead of by the total number of trials. Choice C, 39/2=19.5, divides by 2 for no clear reason connected to the data given. Choice D, 120/39, flips the fraction, dividing the total number of trials by the number of successes instead of the other way around. Estimating probability from a simulation always means dividing the number of successful trials by the total number of trials run.

Question 17

1. Label 5 sections Favorite and 5 sections Not Favorite.
2. Label 6 sections Favorite and 4 sections Not Favorite.
3. Label 4 sections Favorite and 6 sections Not Favorite. (correct answer)
4. Label 1 section Favorite and 9 sections Not Favorite.

Explanation: Since the probability that a song is a favorite is about 0.4, the spinner needs 4 out of its 10 sections labeled Favorite to match that probability, which is exactly what choice C does. Choice A labels 5 sections Favorite, representing a probability of 0.5, which does not match the given 0.4. Choice B labels 6 sections Favorite, representing 0.6, and choice D labels only 1 section Favorite, representing 0.1, both mismatched with the target probability. Getting the individual song probability right is the essential first step before simulating the compound event of choosing 2 favorite songs in a row. Once the spinner is correctly labeled, spinning it twice per trial and checking for two Favorite results would complete the simulation design.

Question 18

1. Use a 12-section spinner; 1 section = March birthday and 11 sections = not March birthday. (correct answer)
2. Flip a coin; heads = March birthday.
3. Use a 10-section spinner; 4 sections = March birthday and 6 sections = not March birthday.
4. Roll a die; 1 = March birthday, 2-6 = not March birthday.

Explanation: The event has a probability of about 1/12, or roughly 0.083, so the best simulation device is one where a single outcome has that same chance - a 12-section spinner with 1 section marked March, matching choice A. A coin gives a 1/2 chance, far too high to represent 1/12. A die gives a 1/6 (about 0.167) chance for any one face, still much higher than 1/12. A 10-section spinner gives at best 1/10 = 0.1 per section, still not as close a match as the 12-section spinner.

Question 19

1. Use a 10-section spinner with 7 sections labeled Win and 3 labeled Lose. Spin 5 times per trial, repeat 10 trials, and estimate P as successes/10.
2. Roll a die 5 times; 1-3 = Win and 4-6 = Lose. Repeat 100 trials and estimate P as successes/100.
3. Flip a coin 5 times; heads = Win. Repeat 100 trials and estimate P as successes/100.
4. Use a 10-section spinner with 3 sections labeled Win and 7 labeled Lose. Spin 5 times per trial, record whether exactly 2 Wins occur, repeat 100 trials, and estimate P as successes/100. (correct answer)

Explanation: A win probability of about 0.3 means the device should have roughly 3 out of 10 equally likely outcomes labeled Win, matching choice D's 10-section spinner with 3 Win sections. Spinning 5 times per trial and checking for exactly 2 Wins simulates the compound event, and running 100 trials gives a stable estimate. Choice A reverses the Win/Lose split (7 Win sections would mean P=0.7, not 0.3) and also uses far too few trials (10). Choices B and C use devices with a 0.5 chance of winning (a die split evenly, or a coin), which don't match the actual 0.3 probability.

Question 20

1. Roll a number cube 3 times per trial (1 = defective); repeat 10 trials; estimate probability.
2. Pick 3 marbles from a bag one time and use that result as the probability.
3. Use a 10-section spinner with 1 section labeled defective and 9 labeled not defective; spin 3 times per trial; repeat 100 trials; count trials with at least one defective; estimate probability. (correct answer)
4. Flip a coin 3 times per trial (heads = defective); repeat 50 trials.

Explanation: With a 10% defect rate, a 10-section spinner with 1 section marked defective matches the probability exactly, so spinning 3 times per trial simulates picking 3 pens, matching choice C. A coin models a 50% chance, far higher than the actual 10% defect rate, and a number cube models about 1/6 (roughly 0.167), also too high. Picking marbles just once doesn't use repeated trials, so it can't produce a reliable probability estimate. Running the spinner for 100 trials, rather than just 10 or 50, gives enough data for a stable estimate.