- Maria wants to estimate the probability that a basketball player makes at least 2 out of 3 free throws. She knows the player has a 60% free throw percentage. Maria decides to simulate this using a random number generator that produces integers from 1 to 10. Which simulation design would be most appropriate for estimating this probability?
Answer: Generate 3 numbers; if at least 2 are between 1-6, count as success. Repeat many times and calculate the proportion of successes.
For a 60% success rate using numbers 1-10, we need 6 out of 10 numbers to represent success (1-6). Since we want to simulate 3 free throws with at least 2 successes, we generate 3 numbers and count success when at least 2 fall in the range 1-6. Choice B uses wrong range (40% instead of 60%). Choice C simulates wrong scenario (6 out of 10 throws). Choice D simulates only 1 throw instead of 3.
- A student runs a simulation to estimate the probability of getting exactly 2 heads when flipping 4 coins. She uses a random number generator producing digits 0-9, where even digits represent heads and odd digits represent tails. After 500 trials, she gets 180 successes. Based on these results, what should she conclude about her simulation?
Answer: The simulation is working correctly; the theoretical probability is 0.375 and her result of 0.36 is very close.
The theoretical probability of exactly 2 heads in 4 flips is C(4,2) × (0.5)4 = 6/16 = 0.375. Her simulation result is 180/500 = 0.36, which is very close. Even and odd digits do represent a fair 50-50 split. Choice B is wrong because simulations give estimates, not exact theoretical values. Choice C is incorrect about the digit assignment. Choice D is wrong because 500 trials is sufficient for a reasonable estimate. - A quality control manager wants to simulate the probability that in a batch of 5 items, at most 1 is defective. Historical data shows 15% of items are defective. She plans to use a spinner divided into 20 equal sections. After running the simulation 1000 times, she should expect approximately how many successful trials?
Answer: 735 successful trials, representing batches with 0 or 1 defective items out of 5
With 15% defect rate, P(0 defective) = (0.85)5 ≈ 0.444 and P(exactly 1 defective) = C(5,1) × (0.15)1 × (0.85)4 ≈ 0.291. So P(at most 1 defective) ≈ 0.444 + 0.291 = 0.735. With 1000 trials, expect about 735 successes. Choice A uses wrong probability calculation. Choice B only counts exactly 1 defective, missing 0 defective cases. Choice C only counts 0 defective cases. - A game designer wants to simulate drawing 3 cards from a standard deck without replacement to find the probability of getting exactly 2 red cards. She decides to use a random number generator producing integers 1-52, where 1-26 represent red cards and 27-52 represent black cards. What is the main flaw in her simulation design?
Answer: The simulation doesn't account for drawing without replacement; once a number is selected, it should be removed from future selections in that trial.
The key flaw is not accounting for drawing without replacement. In her current design, the same card could be 'drawn' multiple times in one trial, which doesn't match the real scenario. Each number selected should be removed from consideration for the remaining draws in that trial. Choice A is wrong because 26 red and 26 black cards correctly represents a standard deck. Choice C would ignore the without-replacement condition. Choice D suggests an inferior method that loses the specific card-drawing context.
- Two students design different simulations for the same probability problem: finding the chance that exactly 2 out of 4 randomly selected students are left-handed (8% of students are left-handed). Student A uses a random number generator 1-100, while Student B uses 1-25. Which statement about their simulation designs is most accurate?
Answer: Both methods are equally valid since 1-8 out of 100 and 1-2 out of 25 both represent 8% accurately in their respective ranges.
Student A uses 1-8 out of 100 numbers (8/100 = 8%) and Student B uses 1-2 out of 25 numbers (2/25 = 8%). Both correctly represent the 8% probability. Choice A incorrectly claims Student B can't represent 8% accurately. Choice B incorrectly claims one method is better than the other. Choice D is wrong because 8% can be simulated accurately with both methods.
- An environmental scientist simulates the probability that exactly 3 out of 5 randomly tested water samples will show pollution levels above the safety threshold. Historical data indicates 25% of samples exceed the threshold. She runs 2000 trials and gets 264 successes. What is the most appropriate way to evaluate her simulation results?
Answer: Recognize that 13.2% seems reasonable for exactly 3 out of 5 successes with 25% individual probability, suggesting the simulation is working correctly.
The theoretical probability of exactly 3 successes out of 5 trials with p=0.25 is C(5,3)(0.25)³(0.75)² = 10×0.015625×0.5625 ≈ 0.088 or about 8.8%. The observed 264/2000 = 13.2% is in the reasonable range considering simulation variability. Choice A is incomplete without doing the theoretical calculation. Choice B misunderstands that 500 would be the expected successes if we wanted 25% overall success rate, not for the specific scenario. Choice D unnecessarily questions the sample size when 2000 trials is quite adequate.
- A teacher designs a simulation to model the probability that in a group of 4 students, at least 3 will pass an exam where the individual pass rate is 70%. She plans to use a deck of cards where hearts and diamonds represent 'pass' and clubs and spades represent 'fail'. After explaining her method, a student points out a problem. What is the most likely issue the student identified?
Answer: The card method only gives a 50% pass rate, not the required 70% pass rate for the simulation.
The main issue is that using hearts/diamonds vs clubs/spades gives a 50% pass rate (26 cards each), not the required 70%. The simulation setup doesn't match the scenario parameters. Choice B raises a valid but secondary concern about replacement vs independence. Choice C misunderstands that 4 cards represents one trial, not the sample size. Choice D incorrectly suggests insufficient cards when there are 26 hearts/diamonds available.
