Finite Mathematics Quiz: Simulation For Probability
19 questions · exam conditions
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Simulation For ProbabilityQuestion 1 of 19

To estimate the probability of a defective part being produced, an engineer designs a simulation. The simulation uses a random integer generator that produces integers from 1 to 200, inclusive. If the generated integer is between 1 and 15, inclusive, the part is recorded as 'defective.' Otherwise, it is recorded as 'not defective.' Based on this simulation design, what is the assumed probability of a part being defective?

0.0700.070
0.0750.075
0.0800.080
0.9250.925
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Finite Mathematics Quiz

Finite Mathematics Quiz: Simulation For Probability

Practice Simulation For Probability in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Simulation For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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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.

All questions

Question 1

To estimate the probability of a defective part being produced, an engineer designs a simulation. The simulation uses a random integer generator that produces integers from 1 to 200, inclusive. If the generated integer is between 1 and 15, inclusive, the part is recorded as 'defective.' Otherwise, it is recorded as 'not defective.' Based on this simulation design, what is the assumed probability of a part being defective?

  1. 0.0700.070
  2. 0.0750.075 (correct answer)
  3. 0.0800.080
  4. 0.9250.925
Explanation: The probability is the ratio of favorable outcomes to the total number of possible outcomes. The total possible outcomes are the integers from 1 to 200, which is 200 numbers. The favorable outcomes (defective) are the integers from 1 to 15, which is 15 numbers. The probability is therefore 15/200=3/40=0.07515/200 = 3/40 = 0.075.

Question 2

A researcher designs a simulation to estimate the probability that in a group of 4 people, at least 2 share the same birth month. Her algorithm generates 4 random integers from 1 to 12 for each trial. After 5000 trials, she observes 2847 successes. She wants to increase the accuracy of her estimate. Which modification would be most effective?

  1. Run multiple sets of 5000 trials and average the results from each set
  2. Change the random number generation to use decimal values between 1.0 and 12.0 instead of integers
  3. Modify the algorithm to generate exactly 12 different birth months and randomly assign people to them
  4. Increase the number of trials to 50,000 while keeping the same algorithm structure (correct answer)
Explanation: When you encounter simulation problems, remember that the goal is to make your estimate as close as possible to the true probability. The accuracy of a simulation estimate depends primarily on sample size - more trials generally mean better estimates due to the Law of Large Numbers. The most direct way to improve accuracy is to increase the number of trials substantially. With 50,000 trials instead of 5,000, you get ten times more data points, which significantly reduces the standard error of your estimate. The current estimate of 2847/5000 = 0.5694 will become more reliable with this larger sample size. Looking at why the other options fall short: Option A (averaging multiple sets of 5000) doesn't actually increase your total sample size - you're just reorganizing the same amount of data. Option B (using decimal values) is problematic because birth months are discrete categories (January, February, etc.), not continuous variables. Using decimals between 1.0 and 12.0 doesn't match the real-world scenario and could introduce artificial precision. Option C fundamentally changes the simulation structure by forcing exactly 12 different months to exist, which doesn't reflect the random nature of the original problem where some months might not appear at all in a trial. Study tip: In simulation questions, always look for the option that increases sample size when you need better accuracy. More trials beat clever reorganization or algorithm modifications when your goal is reducing estimation error.

Question 3

A lottery simulation models the probability of winning by matching exactly 3 out of 6 numbers drawn from 1 to 49. The simulation generates 6 distinct random integers from 1 to 49 for the 'winning numbers' and 6 distinct random integers from 1 to 49 for the 'player numbers', then counts matches. After 100,000 trials, the simulation shows 1,847 wins. The programmer claims this estimate is unreliable because the sample size is too small. What is the most significant issue with this reasoning?

  1. The sample size of 100,000 is actually quite large for this type of simulation and should provide a reliable estimate
  2. The simulation method is incorrect because it doesn't account for the order in which numbers are drawn
  3. The simulation should generate winning numbers once and test many player combinations against those fixed numbers
  4. The issue is not sample size but rather the extremely low theoretical probability making precise estimation difficult even with large samples (correct answer)
Explanation: The theoretical probability of matching exactly 3 out of 6 numbers is very small (approximately 1.8%), so even with 100,000 trials, the number of successes will be small, leading to relatively large confidence intervals around the estimate. The observed 1,847/100,000 ≈ 1.85% is actually quite close to the theoretical value. Choice A misses that small probabilities require very large samples for precision. Choice B is incorrect since lottery order typically doesn't matter. Choice C describes a different valid simulation approach but doesn't address the core issue of estimating small probabilities.

Question 4

A medical researcher simulates the probability that a diagnostic test correctly identifies a disease. The test has a 92% sensitivity (correctly identifies disease when present) and 88% specificity (correctly identifies absence when disease is absent). In her simulation, she assumes 15% of patients have the disease. For each trial, she first determines disease status (15% chance of disease), then determines test result based on the appropriate accuracy rate. After 20,000 trials, she observes 17,024 correct diagnoses. What is the most likely explanation for this result?

  1. The simulation overestimates accuracy because it doesn't properly account for false positives and false negatives
  2. There is likely an error in the algorithm because the overall accuracy should be 90% (average of 92% and 88%)
  3. The simulation is working correctly, as the observed rate of 85.1% matches the expected overall accuracy (correct answer)
  4. The result suggests the 15% disease prevalence assumption is incorrect and should be adjusted
Explanation: When analyzing diagnostic test performance, you need to calculate the overall accuracy by considering both the disease prevalence and the test's sensitivity and specificity rates together. Let's work through the expected accuracy. With 15% disease prevalence, 15% of patients have the disease and 85% don't. For those with disease, the test correctly identifies 92% (sensitivity). For those without disease, the test correctly identifies 88% (specificity). The overall expected accuracy is: (0.15×0.92)+(0.85×0.88)=0.138+0.748=0.886=88.6%(0.15 × 0.92) + (0.85 × 0.88) = 0.138 + 0.748 = 0.886 = 88.6\% With 20,000 trials, we'd expect about 17,720 correct diagnoses. The observed 17,024 correct diagnoses represents 85.1% accuracy, which is reasonably close to the expected 88.6% given normal statistical variation in simulation. Answer A is wrong because the simulation does properly account for false positives and negatives through the sensitivity and specificity rates. Answer B reflects a common misconception—you cannot simply average sensitivity and specificity to get overall accuracy since they apply to different population segments (diseased vs. healthy). The prevalence weights matter crucially. Answer D is incorrect because the observed result is close enough to the expected value that no adjustment to disease prevalence is warranted. Remember: Overall diagnostic accuracy depends on both test performance characteristics AND disease prevalence. You must weight the sensitivity and specificity by their respective population sizes, not simply average them.

Question 5

A simulation estimates the probability that a basketball player makes at least 3 out of 5 free throws. The simulation generates 5 random numbers between 0 and 1 for each trial, counting a 'make' when the number is less than 0.78. After 8000 trials, exactly 6240 trials showed at least 3 makes. If the player's actual free throw percentage in games is 75%, what does this suggest about the simulation model?

  1. The simulation model is appropriate since 0.78 ≈ 0.75, and the observed rate validates the model accuracy
  2. The simulation overestimates the probability because 0.78 > 0.75, leading to more simulated makes than expected (correct answer)
  3. The simulation is flawed because it should use the exact value 0.75 instead of the approximation 0.78
  4. The simulation underestimates the probability because basketball performance involves psychological factors not captured by random numbers
Explanation: The simulation uses a 78% make rate (numbers < 0.78) while the player's actual rate is 75%. This 3 percentage point difference will systematically overestimate the probability of success in each trial, leading to too many simulated makes and an inflated estimate of making at least 3 out of 5. The observed rate of 6240/8000 = 78% is indeed higher than what we'd expect with a 75% shooter. Choice A ignores the meaningful difference between 78% and 75%. Choice C overstates the importance of the approximation. Choice D introduces factors beyond the scope of the basic probability model.

Question 6

A traffic engineer simulates the probability that at least 2 cars arrive at an intersection during a 30-second period. Historical data shows an average of 1.8 cars per 30-second interval. She uses the following algorithm: generate a random number X from an exponential distribution with mean 1.8, then record success if X ≥ 2. After 15,000 simulations, she observes 4,953 successes. Her supervisor argues that this approach incorrectly models the arrival process. What is the primary flaw?

  1. The exponential distribution is inappropriate because car arrivals follow a normal distribution pattern
  2. The algorithm models the time between arrivals rather than the count of arrivals in a fixed time period (correct answer)
  3. The success condition should be X ≥ 1.8 to match the historical average, not X ≥ 2
  4. The simulation should account for traffic light cycles and peak/off-peak variations in arrival rates
Explanation: The engineer is using an exponential distribution (which models time between events) when she should be using a Poisson distribution (which models count of events in a fixed time period). With an average of 1.8 cars per 30 seconds, the number of arrivals follows Poisson(1.8), and she wants P(X ≥ 2) where X is the count of cars. Instead, she's generating inter-arrival times and checking if the first inter-arrival time exceeds 2, which doesn't answer her question. Choice A incorrectly suggests normal distribution. Choice C misunderstands the threshold for 'at least 2'. Choice D introduces complications beyond the basic probability model.

Question 7

Two students run simulations to estimate the probability of getting exactly one head in three coin flips. Student A uses the algorithm: generate three random numbers between 0 and 1, count 'heads' if number < 0.5, record success if exactly one head occurs. Student B uses: generate one random number between 0 and 1, record success if 0.25 ≤ number < 0.625. After 10,000 trials each, Student A observes 3,751 successes and Student B observes 3,748 successes. What can be concluded?

  1. Student A's method is superior because it more accurately models the sequential nature of coin flipping
  2. Student B's method is more efficient and equally valid, since both methods correctly model the same probability distribution (correct answer)
  3. Student B's method is flawed because the interval 0.25 to 0.625 represents the wrong probability
  4. Both methods are flawed because they don't account for the dependence between consecutive coin flips
Explanation: Both methods are mathematically equivalent and valid. The theoretical probability of exactly one head in three flips is 3/8 = 0.375. Student A simulates the actual process (three independent 50% events), while Student B uses the mathematical shortcut of mapping outcomes directly to a single random number. The interval [0.25, 0.625) has length 0.375, correctly representing the target probability. Both observed rates (~37.5%) are very close to the theoretical value. Choice A incorrectly suggests one method is superior. Choice C incorrectly calculates the interval probability. Choice D incorrectly assumes dependence in independent coin flips.

Question 8

A company places one of three different toys (A, B, or C) in each cereal box with equal probability. A simulation is designed to estimate the number of boxes needed to collect all three toys. The simulation uses a standard six-sided die, with rolls 1-2 representing Toy A, 3-4 representing Toy B, and 5-6 representing Toy C. One trial of the simulation produced the following sequence of die rolls: 5, 3, 3, 1, 6, 4, 2. How many boxes were simulated in this trial?

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 7
Explanation: A trial ends when one of each toy has been collected. We map the die rolls to toys: Roll 1 (5) is Toy C. Roll 2 (3) is Toy B. Roll 3 (3) is Toy B (duplicate). Roll 4 (1) is Toy A. At this point, toys A, B, and C have all been collected. The trial finished after the 4th roll, so 4 boxes were simulated.

Question 9

A basketball player makes 72% of her free throws. To simulate the outcomes of her next four free-throw attempts, a student uses a random number generator that produces two-digit integers from 00 to 99. Which of the following assignments correctly models the outcome of a single free-throw attempt?

  1. Let integers 00-71 represent a made free throw and 72-99 represent a missed free throw. (correct answer)
  2. Let integers 00-72 represent a made free throw and 73-99 represent a missed free throw.
  3. Let integers 0-7 represent a made free throw and 8-9 represent a missed free throw.
  4. Let integers 01-72 represent a made free throw and 73-00 (for 100) represent a missed free throw.
Explanation: To model a 72% probability, 72 out of 100 possible outcomes must represent success. Using two-digit integers from 00 to 99 gives 100 possible outcomes. Assigning the 72 integers from 00 to 71 (inclusive) to 'made' correctly represents a 72% chance. The remaining 28 integers (72 to 99) represent 'missed'.

Question 10

A student wants to estimate the probability of passing a 5-question true/false quiz by guessing randomly. A passing grade requires at least 60% correct answers. The student proposes the following simulation: "Flip a fair coin 5 times, corresponding to the 5 questions. Let heads be a correct answer and tails be an incorrect answer. Record the number of heads. Repeat this process for 100 trials. The estimated probability of passing is the number of trials with at least 3 heads, divided by 100." What is the primary flaw, if any, in this simulation design?

  1. The number of trials (100) is too small to provide a meaningful estimate of the probability.
  2. A coin flip incorrectly models guessing, as the chance of guessing a true/false question correctly is not always 50%.
  3. The passing condition is misidentified; 'at least 3 heads' should be 'more than 3 heads'.
  4. The simulation design is sound; there is no primary flaw in its logic for this scenario. (correct answer)
Explanation: The proposed simulation is a valid way to model the scenario. Random guessing on a T/F question implies a 50% chance of being correct, which is correctly modeled by a fair coin flip. A passing grade of 60% on a 5-question quiz is 5×0.60=35 \times 0.60 = 3 questions, so 'at least 3 heads' is the correct success condition. While more trials would increase precision, 100 trials is a valid number for a simulation and not a flaw in the design itself.

Question 11

The following is a line of random digits: 39899 04005 1...

A traffic light operates on a 100-second cycle. It is red for 40 seconds, green for 50 seconds, and yellow for 10 seconds. To simulate the color of the light encountered by cars arriving at random moments, two-digit numbers from 00 to 99 are used. The assignment is: 00-39 for red, 40-89 for green, and 90-99 for yellow. Using the line of random digits from the provided passage, what is the sequence of light colors for the first five simulated cars?

  1. Red, Green, Yellow, Green, Red (correct answer)
  2. Red, Green, Green, Red, Red
  3. Green, Red, Yellow, Green, Green
  4. Red, Green, Yellow, Red, Red
Explanation: We group the random digits into pairs: 39, 89, 90, 40, 05. Then we apply the given assignment: 39 is in the range 00-39, so it's Red. 89 is in 40-89, so it's Green. 90 is in 90-99, so it's Yellow. 40 is in 40-89, so it's Green. 05 is in 00-39, so it's Red. The resulting sequence is Red, Green, Yellow, Green, Red.

Question 12

In a certain city, 60% of days are sunny and 40% are cloudy, with each day's weather being independent of the others. A researcher wants to simulate the number of days until she observes two consecutive sunny days. She uses two-digit random numbers from 00-99. Which procedure correctly simulates one trial of this experiment?

  1. Let 00-59 be 'sunny' and 60-99 be 'cloudy'. Generate a list of 10 random numbers and count how many pairs of consecutive sunny days appear.
  2. Let 00-59 be 'sunny' and 60-99 be 'cloudy'. Generate numbers until any two numbers in the 'sunny' range have appeared. Count the total numbers generated.
  3. Let 00-59 be 'sunny' and 60-99 be 'cloudy'. Generate numbers sequentially until two consecutive numbers are in the 00-59 range. The count of numbers generated is the result of the trial. (correct answer)
  4. Let 0-5 be 'sunny' and 6-9 be 'cloudy'. Generate numbers sequentially until two consecutive numbers are in the 0-5 range. The count of numbers generated is the result of the trial.
Explanation: The goal is to find the waiting time for a specific event ('two consecutive sunny days'). The simulation must generate outcomes sequentially and stop as soon as the condition is met. Assigning 60 out of 100 numbers (00-59) to 'sunny' correctly models the 60% probability. The trial ends only when two consecutive outcomes are sunny. The total number of days (numbers generated) until this occurs is the result of one trial.

Question 13

To estimate the area of an irregular lake on a map, a researcher overlays a 20 km by 20 km square grid. A computer simulation then randomly generates 10,000 coordinate points (x,y)(x, y) within this square. The simulation finds that 2,350 of these points fall inside the boundaries of the lake. Based on this simulation, what is the approximate area of the lake?

  1. 23.5 km2^2
  2. 0.235 km2^2
  3. 2350 km2^2
  4. 94 km2^2 (correct answer)
Explanation: This question tests Monte Carlo simulation, a method for estimating areas or probabilities using random sampling. When you see random points being generated within a known area to estimate an unknown area, you're dealing with proportional reasoning based on hit rates. The key insight is that the ratio of points inside the lake to total points should equal the ratio of the lake's area to the total grid area. You have a 20 km × 20 km grid, so the total area is 20×20=40020 \times 20 = 400 km². Out of 10,000 random points, 2,350 fell inside the lake. Setting up the proportion: points in laketotal points=lake areagrid area\frac{\text{points in lake}}{\text{total points}} = \frac{\text{lake area}}{\text{grid area}} 2,35010,000=lake area400\frac{2,350}{10,000} = \frac{\text{lake area}}{400} 0.235=lake area4000.235 = \frac{\text{lake area}}{400} lake area=0.235×400=94 km2\text{lake area} = 0.235 \times 400 = 94 \text{ km}^2 Answer choice A (23.5 km²) represents the common error of forgetting to multiply by the total grid area—you'd get this if you mistakenly thought the proportion itself was the area. Answer B (0.235 km²) is just the raw proportion without any area conversion. Answer C (2,350 km²) comes from treating the number of points as the area directly, ignoring the scaling factor entirely. Remember: in Monte Carlo problems, always set up the proportion between hit rates and areas, then solve for the unknown area using the known total area of your sampling region.

Question 14

A fair six-sided die is rolled 60 times to simulate a game. The theoretical expected number of times the outcome '4' will occur is 10. In the actual simulation, the outcome '4' occurs 13 times. Which statement best explains the relationship between the theoretical expectation and the simulation outcome?

  1. The die must not be fair since the observed frequency (13) deviates from the expected frequency (10).
  2. The theoretical probability is incorrect since the simulation shows the true probability of rolling a '4' is 13/6013/60.
  3. The discrepancy is due to random variation inherent in probabilistic experiments, and different trials would yield different results. (correct answer)
  4. The simulation was not run long enough since the Law of Large Numbers would eventually make the observed frequency closer to 10.
Explanation: Theoretical probability describes long-term averages, not exact outcomes for a specific, finite number of trials. Random variation means that in any given experiment, the observed frequency of an event is likely to differ from its theoretical expectation. A result of 13 when 10 is expected is a plausible outcome due to this randomness. A different simulation would likely produce another different, but also plausible, result.

Question 15

An urn contains 5 red and 3 blue marbles. A simulation is designed to estimate the probability that the second marble drawn is red, given that the first marble drawn was red (P(2nd is R∣1st is R)P(\text{2nd is R} | \text{1st is R})). Marbles are drawn without replacement. Which procedure correctly estimates this conditional probability?

  1. Simulate 1000 trials of drawing two marbles. Count the number of trials where the second marble is red and divide by 1000.
  2. Simulate 1000 trials of drawing two marbles. Count the number of trials where both marbles are red and divide by 1000.
  3. Simulate many trials. Among only the trials where the first draw was red, calculate the proportion where the second draw was also red. (correct answer)
  4. Simulate many trials. Among only the trials where the second draw was red, calculate the proportion where the first draw was also red.
Explanation: To estimate a conditional probability P(A∣B)P(A|B), we must simulate the entire process but then restrict our analysis to the subset of outcomes where the condition B has occurred. In this case, the condition is 'the first marble drawn was red'. Therefore, we perform many trials, but we only consider the trials that began with a red draw. Within that specific subset of trials, we find the proportion that had a red marble as the second draw. This correctly simulates the conditional probability.

Question 16

A manufacturer knows that 8% of its products are defective. Two students are asked to simulate the process of selecting 10 products from the assembly line. Student A runs the simulation for 50 trials. Student B runs the simulation for 500 trials. Let PAP_A be the proportion of trials with at least one defective product found by Student A, and PBP_B be the proportion found by Student B. Let PTP_T be the true theoretical probability. Which statement is most likely to be true regarding the results?

  1. PAP_A is more likely to be closer to PTP_T than PBP_B is.
  2. PBP_B is more likely to be closer to PTP_T than PAP_A is. (correct answer)
  3. It is certain that the value of PBP_B will be closer to PTP_T than the value of PAP_A.
  4. It is certain that PAP_A and PBP_B will be equal to each other, but not necessarily equal to PTP_T.
Explanation: The Law of Large Numbers states that as the number of trials in a probability experiment increases, the experimental probability will tend to get closer to the theoretical probability. Since Student B conducted more trials (500) than Student A (50), Student B's result (PBP_B) is more likely to be a better estimate of the theoretical probability (PTP_T) than Student A's result (PAP_A). It is not a certainty, as random chance could lead to a less-representative sample in the larger simulation, but it is the most likely outcome.

Question 17

A quality control manager simulates the probability that exactly 2 out of 5 randomly selected items are defective. In her simulation, she generates 5 random numbers between 0 and 1 for each trial, counting a number as 'defective' if it's less than 0.15. She runs 2000 trials and observes that exactly 2 items were defective in 142 trials. If she had instead defined 'defective' as numbers greater than 0.85, approximately how many trials would have shown exactly 2 defective items?

  1. Approximately 142 trials, because the probability of exactly 2 successes remains the same (correct answer)
  2. Approximately 284 trials, because the new method doubles the success rate
  3. Approximately 71 trials, because the new method halves the number of successes
  4. Approximately 1858 trials, because most trials would now show 2 or fewer defects
Explanation: Both conditions (< 0.15 and > 0.85) represent exactly 15% of the random number range, so the probability of any single item being defective remains 0.15. The binomial probability of exactly 2 out of 5 items being defective is the same in both cases. Therefore, approximately 142 trials would show exactly 2 defective items. Choice B incorrectly assumes the rate doubles. Choice C incorrectly assumes the rate halves. Choice D confuses 'exactly 2' with 'at most 2' defective items.

Question 18

A researcher wants to estimate the probability that a randomly selected student passes both a math test and a science test. She runs a simulation with 1000 trials where each trial generates two random numbers between 0 and 1. If the first number is less than 0.72 (representing passing math) AND the second number is less than 0.68 (representing passing science), she counts it as a success. After running the simulation, she observes 487 successes. Based on this simulation, what should her estimated probability be, and how does it compare to the theoretical probability?

  1. Estimated: 0.487, Theoretical: 0.490, so the simulation slightly underestimated the true probability (correct answer)
  2. Estimated: 0.487, Theoretical: 0.700, so the simulation significantly underestimated the true probability
  3. Estimated: 0.513, Theoretical: 0.490, so the simulation slightly overestimated the true probability
  4. Estimated: 0.487, Theoretical: 0.140, so the simulation significantly overestimated the true probability
Explanation: The estimated probability from simulation is 487/1000 = 0.487. The theoretical probability of both independent events occurring is 0.72 × 0.68 = 0.4896 ≈ 0.490. The simulation result is very close to the theoretical value, slightly underestimating it. Choice B incorrectly adds the probabilities (0.72 + 0.68 = 1.40, but capped at 1.0 gives 0.70). Choice C uses the wrong simulation result (1000 - 487 = 513). Choice D incorrectly subtracts probabilities.

Question 19

A game designer simulates a dice game where players roll two dice and win if the sum is 7 or 11. She wants to estimate the probability using two different simulation approaches. Method 1: Generate two random integers from 1 to 6, check if sum equals 7 or 11. Method 2: Generate one random number from 0 to 1, win if 0 ≤ number < 2/9. After running each method 12,000 times, Method 1 shows 2,667 wins and Method 2 shows 2,664 wins. What conclusion should she draw?

  1. Method 1 is more accurate because it directly simulates the physical dice rolling process
  2. Method 2 is flawed because the probability interval 2/9 is incorrect for this game
  3. Both methods are mathematically equivalent and the small difference in results is due to random variation (correct answer)
  4. Method 2 is superior because it's computationally more efficient while maintaining the same accuracy
Explanation: The correct probability for rolling a sum of 7 or 11 with two dice is 8/36 = 2/9 ≈ 0.222. Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways. Sum of 11: (5,6), (6,5) = 2 ways. Total: 8 ways out of 36 possible outcomes = 2/9. Method 2 correctly uses this probability. Both observed results (2667/12000 ≈ 0.222 and 2664/12000 ≈ 0.222) are very close to the theoretical value, confirming both methods are equivalent. The small difference is normal sampling variation.