Question 1190703: Residents in an inner-city area are concerned about drug dealers entering their neighborhood. Over the past 14 nights, they have taken turns watching the street from a darkened apartment. Drug deals seem to take place randomly at various times and locations on the street and average about three per night. The residents of this street contacted the local police, who informed them that they do not have sufficient resources to set up surveillance. The police suggested videotaping the activity on the street, and if the residents are able to capture five or more drug deals on tape, the police will take action. Unfortunately, none of the residents on this street owns a video camera and, hence, they would have to rent the equipment. Inquiries at the local dealers indicated that the best available rate for renting a video camera is $75 for the first night and $40 for each additional night. To obtain this rate, the residents must sign up in advance for a specified number of nights. The residents hold a neighborhood meeting and invite you to help them decide on the length of the rental period. Because it is difficult for them to pay the rental fees, they want to know the probability of taping at least five drug deals on a given number of nights of videotaping. a. Which of the probability distributions you have studied might be helpful here? b. What assumption(s) would you have to make? c. If the residents tape for two nights, what is the probability they will film at least five drug deals? d. For how many nights must the camera be rented so that there is at least .90 probability that five or more drug deals will be taped?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this problem:
**a. Which probability distribution?**
The **Poisson distribution** is appropriate here. The Poisson distribution models the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known average rate and independently of the time since the last event. In this case, the "events" are drug deals, the "interval" is nights, and the average rate is 3 deals per night.
**b. Assumptions:**
To use the Poisson distribution, we need to make the following assumptions:
1. **Independence:** The number of drug deals on one night is independent of the number of deals on any other night.
2. **Constant Rate:** The average rate of drug deals is constant at 3 per night. This might not be entirely true in reality (it could be higher on weekends, for example), but it's the assumption we're working with.
3. **Randomness:** The drug deals occur randomly.
**c. Probability of at least five deals in two nights:**
1. **Adjust the average rate:** Since the average is 3 deals per night, over two nights, the average would be 3 deals/night * 2 nights = 6 deals. This becomes our new λ (lambda) for the Poisson distribution.
2. **Poisson Formula:** The probability of *k* events occurring is: P(k) = (e^(-λ) * λ^k) / k!
3. **Calculate P(k ≥ 5):** We want the probability of *at least* five deals, which means 5, 6, 7, and so on. It's easier to calculate the complement (0, 1, 2, 3, or 4 deals) and subtract from 1:
P(k ≥ 5) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4)]
Calculate each probability using the Poisson formula (with λ = 6) and sum them.
P(k ≥ 5) = 1 - [(e^-6 * 6^0)/1! + (e^-6 * 6^1)/1! + (e^-6 * 6^2)/2! + (e^-6 * 6^3)/3! + (e^-6 * 6^4)/4!]
P(k ≥ 5) ≈ 1 - [0.0025 + 0.0149 + 0.0446 + 0.0892 + 0.1339]
P(k ≥ 5) ≈ 1 - 0.2851
P(k ≥ 5) ≈ 0.7149
**d. Nights needed for at least 0.90 probability:**
This requires a bit of trial and error. We need to find the smallest number of nights (n) such that the probability of at least 5 deals is greater than or equal to 0.90.
1. **Adjust the average rate:** The average rate over *n* nights will be 3n.
2. **Calculate probabilities:** For each value of *n*, calculate P(k ≥ 5) using the Poisson distribution with λ = 3n, as we did in part (c).
3. **Stop when the probability is reached:** Continue increasing *n* until P(k ≥ 5) reaches at least 0.90.
Here's a table to help illustrate the process. The calculations below are rounded to 4 decimal places.
| Nights (n) | Average Rate (λ = 3n) | P(k≥5) |
|---|---|---|
| 2 | 6 | 0.7149 |
| 3 | 9 | 0.9112 |
Therefore, the residents need to rent the camera for *at least 3 nights* to have a probability greater than 0.90 of filming at least 5 drug deals.
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