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The problem given in the image refers to calculating probabilities for rainfall flooding based on a **return period of 10 years**. This means that the probability of flooding in any given year is \( p = \frac{1}{10} = 0.1 \), and the probability of no flooding in any given year is \( q = 1 - p = 0.9 \).

We will now solve each part of the problem using this information.

---

### (a) Probability that the sewers will be flooded for the first time in the third year

This scenario follows a **geometric distribution**, where the probability of the first success (flood) occurs after two failures (no flood in the first two years), followed by a flood in the third year.

The probability is calculated as:
\[
P(\text{flood in third year}) = q^2 \cdot p = (0.9)^2 \cdot 0.1 = 0.81 \cdot 0.1 = 0.081
\]

So, the probability is **0.081**.

---

### (b) Probability of flooding within the first 3 years

The probability of at least one flood in 3 years can be found using the complement rule. First, we calculate the probability of **no flood** in the first 3 years, which is \( q^3 \), and subtract this from 1.

\[
P(\text{no flood in 3 years}) = q^3 = (0.9)^3 = 0.729
\]
\[
P(\text{at least one flood}) = 1 - P(\text{no flood}) = 1 - 0.729 = 0.271
\]

So, the probability of flooding within the first 3 years is **0.271**.

---

### (c) Probability of flooding in 3 of the first 5 years

Here, we use the **binomial distribution** to calculate the probability of exactly 3 floods in 5 years, where \( p = 0.1 \) and \( q = 0.9 \). The binomial probability formula is:

\[
P(X = 3) = \binom{5}{3} \cdot p^3 \cdot q^2
\]

Where \( \binom{5}{3} \) is the binomial coefficient. First, let's compute that:

\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = 10
\]

Now, we plug in the values:

\[
P(X = 3) = 10 \cdot (0.1)^3 \cdot (0.9)^2 = 10 \cdot 0.001 \cdot 0.81 = 0.0081
\]

So, the probability of flooding in 3 of the first 5 years is **0.0081**.

---

### (d) Probability of only one flood within 3 years

Again, using the **binomial distribution**, we calculate the probability of exactly one flood in 3 years. The formula is:

\[
P(X = 1) = \binom{3}{1} \cdot p^1 \cdot q^2
\]
Where \( \binom{3}{1} = 3 \).

\[
P(X = 1) = 3 \cdot (0.1)^1 \cdot (0.9)^2 = 3 \cdot 0.1 \cdot 0.81 = 0.243
\]

So, the probability of only one flood within 3 years is **0.243**.

---

### Summary of Results:
- (a) Probability of flooding for the first time in the third year: **0.081**
- (b) Probability of flooding within the first 3 years: **0.271**
- (c) Probability of flooding in 3 of the first 5 years: **0.0081**
- (d) Probability of only one flood within 3 years: **0.243**

---

Do you need further details or explanations? Here are five questions to expand on the problem:

1. How would the probabilities change if the return period was 20 years instead of 10?
2. Can you explain how the geometric distribution is derived for this type of problem?
3. What if we were asked to calculate the probability of no floods in the first 5 years?
4. How do we generalize this approach for n years instead of just 3 or 5?
5. Can you compare the binomial distribution to the Poisson distribution in the context of rare events?

**Tip:** Always check the return period carefully when solving flooding probability problems, as it affects all subsequent probability calculations.

What is the probability that the sewers in a city will flood based on different conditions, considering a return period of 10 years?

Learn how to calculate the probability of flooding for sewers in a city based on different scenarios, including flooding for the first time in a specific year, within 3 years, or over multiple years. This problem uses key concepts like geometric distribution, binomial distribution, and complement rule. Suitable for advanced high school students in Grades 11-12.

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