Probability calculations for customer influx, guitar defect, and Halloween house visits
Probability calculations for customer influx, guitar defect, and Halloween house visits
A local drugstore owner knows that, on average, 20 people enter his store each hour, assuming that the number of people that come into the store follows a Poisson process. Find the probability that at least 10 people will come to the store in 30 minutes. Find the probability that at least 5 people will come to the store in 15 minutes.
A store has 20 guitars in stock, but 3 are defective. Claire buys 5 guitars from this lot. Find the probability that Claire bought 2 defective guitars. Suppose that 3% of a total of 5000 guitars are defective. Claire buys 10 guitars from this lot. Find the probability that Claire bought 2 defective guitars from this lot.
It is estimated that 5 out of every 8 houses give away candy for Halloween. If we want to receive candy from 5 different houses, what is the probability that we will need to visit at least 10 of them? If you plan to visit 26 houses, what is the probability that you will get candy in at least 15 of them?
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The following analysis explores various probability scenarios involving Poisson processes, hypergeometric, and binomial distributions, each relevant to real-world contexts such as customer flow, product defect rates, and social behaviors during Halloween. These statistical models provide insight into the likelihood of specific events, enabling businesses and individuals to make informed decisions based on probabilistic outcomes.
Poisson Distribution and Customer Traffic
The Poisson distribution models the number of events (e.g., customer arrivals) occurring within a fixed interval, assuming these events happen independently at a constant average rate. In this case, the store expects an average of 20 customers per hour. To find the probability of at least 10 customers arriving within 30 minutes, we first determine the expected number of arrivals during that period. Since the average is 20 per hour, the expected number in 30 minutes (half an hour) is λ = 20 / 2 = 10 customers.
The probability that at least k events occur in a Poisson distribution is given by:
P(X ≥ k) = 1 - P(X ≤ k - 1) = 1 - ∑_{i=0}^{k-1} (e^{-λ} λ^i) / i!
For part (a), k=10, λ=10; thus, the probability of at least 10 customers in 30 minutes is:

P(X ≥ 10) = 1 - P(X ≤ 9) = 1 - ∑_{i=0}^{9} (e^{-10} 10^i) / i!
Similarly, for part (b), in 15 minutes, the expected number of customers is λ = 20 / 4 = 5, and we want the probability that at least 5 customers arrive:
P(X ≥ 5) = 1 - P(X ≤ 4) = 1 - ∑_{i=0}^{4} (e^{-5} 5^i) / i!
These computations often involve tabulated Poisson probabilities or computational tools for precise results, but conceptually, they reflect the likelihood of observing a minimum number of visitors during given time frames based on the Poisson model.
Hypergeometric Distribution and Product Defect Rates
The hypergeometric distribution applies when sampling without replacement from a finite population. In the first scenario, there are 20 guitars, 3 of which are defective, and Claire randomly selects 5 guitars. The probability that she ends up with exactly 2 defective guitars is:
P(X=2) = (C(3,2) * C(17,3)) / C(20,5)
Where C(n,k) denotes combinations. This calculation considers the number of ways to select 2 defective guitars from 3 defective ones, combined with selecting 3 non-defective guitars from 17 non-defective, divided by total ways to select any 5 guitars.
In the second scenario, with a defect rate of 3% in a large batch of 5000 guitars, Claire buys 10 guitars. Assuming the defectiveness follows a binomial distribution (which approximates hypergeometric for large populations), the probability she purchases exactly 2 defective guitars is:
P(X=2) = C(10,2) * (0.03)^2 * (0.97)^8
This binomial probability provides a practical estimate assuming each guitar's defectiveness is independent, and the defect rate remains consistent in the population.
Binomial Distribution and Halloween Social Behavior
Based on the estimate that 5 out of 8 houses give away candy, the probability that a randomly selected house offers candy is p=5/8=0.625. To determine the probability of visiting at least 10 houses before receiving candy from 5 different houses, one can model the number of houses visited until the fifth "success" (candy-giving house) using a negative binomial distribution:

P (visiting at least 10 houses to get 5 successes) = P (Number of trials ≥ 10 to achieve 5 successes).
This corresponds to the sum over all outcomes where the number of houses visited is at least 10, considering the negative binomial distribution parameters. Alternatively, considering the problem as a sequence of Bernoulli trials, the probability of needing at least 10 visits is computed by summing the probabilities for exactly 10, 11, ... visits until the fifth success.
For the second part, expecting at least 15 successful candy receipts out of 26 visits, the binomial distribution directly applies, with p=0.625. The probability of getting at least 15 candies out of 26 houses is:
P(X ≥ 15) = ∑_{k=15}^{26} C(26,k) p^k (1 - p)^{26 - k}
This probability indicates how likely it is to receive candy from at least half of the houses visited, given the success rate per house.
Practical Implications and Applications
Understanding these probability models aids in strategic decision-making across various fields. Retailers can predict customer traffic to optimize staffing; manufacturers assess defect rates to improve quality control; and social planners estimate community engagement levels in events like Halloween, ensuring resource allocation matches expected participation. These models exemplify the utility of probabilistic thinking in managing uncertainty and planning effectively.
Conclusion
In conclusion, analyzing probabilistic scenarios using Poisson, hypergeometric, and binomial distributions provides valuable insights into real-world events. Mastery of these models enhances decision-making accuracy, supports risk assessment, and fosters a data-driven approach to problem-solving across diverse domains.
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