It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
i) exactly 5 rats (ii) more than 5 rats (iii) between 5 and 7 rats, inclusive.
Given `e^(-5) =0.0067`.

Answer

To solve the problem, we will use the Poisson distribution formula, which is given by:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

where \( \lambda \) is the average rate (mean), \( k \) is the number of occurrences, and \( e \) is Euler's number (approximately equal to 2.71828).

Given that the average number of rats per bungalow is \( \lambda = 5 \) and \( e^{-5} = 0.0067 \), we will calculate the required probabilities step by step.

### i) Probability of exactly 5 rats

To find the probability of having exactly 5 rats, we set \( k = 5 \):

\[ P(X = 5) = \frac{e^{-5} \cdot 5^5}{5!} \]

Calculating \( 5^5 \) and \( 5! \):

- \( 5^5 = 3125 \)
- \( 5! = 120 \)

Now substituting these values into the formula:

\[ P(X = 5) = 0.0067 \cdot \frac{3125}{120} \]

Calculating the fraction:

\[ \frac{3125}{120} = 26.04167 \]

Now, substituting back:

\[ P(X = 5) = 0.0067 \cdot 26.04167 \approx 0.174 \]

### ii) Probability of more than 5 rats

To find the probability of having more than 5 rats, we can use the complement rule:

\[ P(X > 5) = 1 - P(X \leq 5) \]

First, we need to calculate \( P(X \leq 5) \):

\[ P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \]

Calculating each term:

1. \( P(X = 0) = \frac{e^{-5} \cdot 5^0}{0!} = 0.0067 \cdot 1 = 0.0067 \)
2. \( P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 0.0067 \cdot 5 = 0.0335 \)
3. \( P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = 0.0067 \cdot \frac{25}{2} = 0.08375 \)
4. \( P(X = 3) = \frac{e^{-5} \cdot 5^3}{3!} = 0.0067 \cdot \frac{125}{6} \approx 0.13958 \)
5. \( P(X = 4) = \frac{e^{-5} \cdot 5^4}{4!} = 0.0067 \cdot \frac{625}{24} \approx 0.174 \) (already calculated)
6. \( P(X = 5) = 0.174 \)

Now summing these probabilities:

\[ P(X \leq 5) = 0.0067 + 0.0335 + 0.08375 + 0.13958 + 0.174 \approx 0.437 \]

Now, we can find \( P(X > 5) \):

\[ P(X > 5) = 1 - P(X \leq 5) = 1 - 0.437 \approx 0.563 \]

### iii) Probability of between 5 and 7 rats (inclusive)

To find the probability of having between 5 and 7 rats, we calculate:

\[ P(5 \leq X \leq 7) = P(X = 5) + P(X = 6) + P(X = 7) \]

We already have \( P(X = 5) = 0.174 \).

Now calculating \( P(X = 6) \) and \( P(X = 7) \):

1. \( P(X = 6) = \frac{e^{-5} \cdot 5^6}{6!} = 0.0067 \cdot \frac{15625}{720} \approx 0.145 \)
2. \( P(X = 7) = \frac{e^{-5} \cdot 5^7}{7!} = 0.0067 \cdot \frac{78125}{5040} \approx 0.103 \)

Now summing these probabilities:

\[ P(5 \leq X \leq 7) = 0.174 + 0.145 + 0.103 \approx 0.422 \]

### Final Answers
- i) \( P(X = 5) \approx 0.174 \)
- ii) \( P(X > 5) \approx 0.563 \)
- iii) \( P(5 \leq X \leq 7) \approx 0.422 \)