(1 point) The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best rate with 80% of its flights arriving on time. A test is conducted by randomly selecting 17
 Southwest flights and observing whether they arrive on time.
 Find the probability that at least 4
 flights arrive late.

This problem involves a binomial distribution since we are dealing with a fixed number of trials (17 flights) and a fixed probability of success (arriving on time). The probability of a flight arriving on time is given as 80%, which means the probability of a flight arriving late is 20%.

We want to find the probability that at least 4 flights arrive late. This can be written as \( P(X \geq 4) \), where \( X \) is the number of late flights, which follows a binomial distribution:

\[
X \sim \text{Binomial}(n = 17, p = 0.2)
\]

The binomial probability mass function is:

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

Where:
- \( n = 17 \) (number of flights),
- \( p = 0.2 \) (probability of a flight arriving late),
- \( k \) is the number of late flights.

To find \( P(X \geq 4) \), we can use the complement rule:

\[
P(X \geq 4) = 1 - P(X \leq 3)
\]

So, we need to calculate \( P(X = 0) \), \( P(X = 1) \), \( P(X = 2) \), and \( P(X = 3) \), and then subtract their sum from 1.

### Step-by-Step Calculation:

1. **Calculate individual probabilities** for \( X = 0 \), \( X = 1 \), \( X = 2 \), and \( X = 3 \).

   \[
   P(X = 0) = \binom{17}{0} (0.2)^0 (0.8)^{17} = 0.0387
   \]
   
   \[
   P(X = 1) = \binom{17}{1} (0.2)^1 (0.8)^{16} = 0.1645
   \]
   
   \[
   P(X = 2) = \binom{17}{2} (0.2)^2 (0.8)^{15} = 0.3085
   \]
   
   \[
   P(X = 3) = \binom{17}{3} (0.2)^3 (0.8)^{14} = 0.2701
   \]

2. **Sum these probabilities**:

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

   \[
   P(X \leq 3) = 0.0387 + 0.1645 + 0.3085 + 0.2701 = 0.7818
   \]

3. **Find the complement**:

   \[
   P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.7818 = 0.2182
   \]

### Final Answer:
The probability that at least 4 flights arrive late is approximately **0.2182** (or 21.82%).

---

Do you want more details on any of the steps, or have any questions?

Here are 5 related questions:
1. What is the probability that exactly 5 flights arrive late?
2. How would the probability change if 10% of flights were late?
3. What is the expected number of late flights for Southwest Air?
4. How does the binomial distribution change if we increase the number of trials (flights)?
5. What are the odds of having all 17 flights arrive on time?

**Tip:** Using the complement rule is often simpler when calculating "at least" probabilities in binomial distributions!

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best rate with 80% of its flights arriving on time. A test is conducted by randomly selecting 17 Southwest flights and observing whether they arrive on time. Find the probability that at least 4 flights arrive late.

Learn how to solve a binomial probability problem involving Southwest Airlines' on-time performance. This example covers the probability that at least 4 flights arrive late, using key concepts like binomial distribution and the complement rule. Suitable for Grades 10-12.

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