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.

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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.