The problem involves determining the probability of at least 3 flights arriving late out of 14 flights, given that 80% of Southwest flights arrive on time. Here’s how to calculate it step by step:

Step 1: Understand the problem

• Probability of being on time: $p = 0.8$
• Probability of being late: $1 - p = 0.2$
• Number of flights: $n = 14$
• We want the probability of at least 3 flights arriving late, i.e., $P(X \geq 3)$, where $X$ is the number of late flights.

Step 2: Use the complement rule

The complement of $P(X \geq 3)$ is $P(X \leq 2)$. Thus: $P(X \geq 3) = 1 - P(X \leq 2)$ This means we need to calculate $P(X = 0) + P(X = 1) + P(X = 2)$, and subtract it from 1.

Step 3: Binomial probability formula

The probability of exactly $k$ successes (late flights) in $n$ trials is given by the binomial formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step 4: Compute probabilities for $X = 0, 1, 2$

For $X = 0$:

$P(X = 0) = \binom{14}{0} (0.2)^0 (0.8)^{14} = 1 \cdot 1 \cdot (0.8)^{14} \approx 0.0439$

For $X = 1$:

$P(X = 1) = \binom{14}{1} (0.2)^1 (0.8)^{13} = 14 \cdot 0.2 \cdot (0.8)^{13} \approx 0.153$

For $X = 2$:

$P(X = 2) = \binom{14}{2} (0.2)^2 (0.8)^{12} = \frac{14 \cdot 13}{2} \cdot (0.2)^2 \cdot (0.8)^{12} \approx 0.264$

Step 5: Add probabilities for $X = 0, 1, 2$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0439 + 0.153 + 0.264 = 0.4609$

Step 6: Subtract from 1

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.4609 = 0.5391$

Final Answer:

The probability that at least 3 flights arrive late is approximately 0.539.

Let me know if you want a deeper breakdown or have questions about the calculations!

Related Questions:

1. What is the probability of exactly 4 flights being late?
2. How do you calculate $P(X \leq 5)$?
3. What is the expected number of late flights?
4. What is the standard deviation of the number of late flights?
5. How would the result change if $p = 0.9$?

Tip:

When solving binomial probability problems, using a calculator or software to handle powers and factorials can save a lot of time and reduce errors!