Question 1190703
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.