This problem follows a binomial distribution with \( n = 17 \) properties and a success probability \( p = 0.3 \). We want the probability of selling more than 4 properties, which is \( P(X > 4) \).

First, calculate \( P(X \leq 4) \) using the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Then subtract from 1 to find \( P(X > 4) \). This requires calculating \( P(X = 0) \) through \( P(X = 4) \):

1. \( P(X = 0) \)
2. \( P(X = 1) \)
3. \( P(X = 2) \)
4. \( P(X = 3) \)
5. \( P(X = 4) \)

Summing these and subtracting from 1:

\[
P(X > 4) = 1 - \sum_{k=0}^{4} P(X = k)
\]

Computing this, we find:

\[
P(X > 4) \approx 0.7477
\]

Thus, the probability is approximately 0.7477.

Let's break down this binomial distribution problem to find \( P(X > 4) \).

Given:
- \( n = 17 \) (the number of trials),
- \( p = 0.3 \) (the probability of success in a single trial),
- We want to calculate \( P(X > 4) \), where \( X \) is the number of successes.

### Step 1: Calculate \( P(X \leq 4) \)
Using the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
we need to compute \( P(X = 0) \) through \( P(X = 4) \), then sum these probabilities to get \( P(X \leq 4) \).

#### Calculations:

1. **\( P(X = 0) \)**:
\[
P(X = 0) = \binom{17}{0} (0.3)^0 (0.7)^{17} = 1 \cdot 1 \cdot (0.7)^{17} \approx 0.00526
\]

2. **\( P(X = 1) \)**:
\[
P(X = 1) = \binom{17}{1} (0.3)^1 (0.7)^{16} = 17 \cdot 0.3 \cdot (0.7)^{16} \approx 0.03828
\]

3. **\( P(X = 2) \)**:
\[
P(X = 2) = \binom{17}{2} (0.3)^2 (0.7)^{15} = 136 \cdot (0.3)^2 \cdot (0.7)^{15} \approx 0.11798
\]

4. **\( P(X = 3) \)**:
\[
P(X = 3) = \binom{17}{3} (0.3)^3 (0.7)^{14} = 680 \cdot (0.3)^3 \cdot (0.7)^{14} \approx 0.21026
\]

5. **\( P(X = 4) \)**:
\[
P(X = 4) = \binom{17}{4} (0.3)^4 (0.7)^{13} = 2380 \cdot (0.3)^4 \cdot (0.7)^{13} \approx 0.24578
\]

### Step 2: Sum these probabilities

\[
P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\]
\[
P(X \leq 4) \approx 0.00526 + 0.03828 + 0.11798 + 0.21026 + 0.24578 = 0.61756
\]

### Step 3: Calculate \( P(X > 4) \)
\[
P(X > 4) = 1 - P(X \leq 4) = 1 - 0.61756 = 0.38244
\]

So, the probability of selling more than 4 properties is approximately **0.3824**.

---

### Relative Questions:

1. What happens to the probability \( P(X > 4) \) if the success probability \( p \) is increased?
2. How does increasing \( n \) to 20 change the probability \( P(X > 4) \)?
3. What is the variance of this binomial distribution with \( n = 17 \) and \( p = 0.3 \)?
4. How would the probability change if we are interested in \( P(X > 7) \) instead of \( P(X > 4) \)?
5. What is the expected value \( E(X) \) for this distribution?

### Tip:
When dealing with binomial probabilities, always check if the cumulative distribution tables or software tools are available to ease manual computations, especially for large \( n \) values.

This problem follows a binomial distribution with n = 17 properties and a success probability p = 0.3. We want the probability of selling more than 4 properties, which is P(X > 4).

Learn how to calculate the probability of selling more than 4 properties using the binomial distribution. This problem demonstrates the use of binomial probability formulas with n = 17 and a success probability of p = 0.3. Suitable for students in Grades 10-12, it covers concepts like cumulative probability and the binomial theorem.

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