From experience, an airline knows that only 70% of the passengers booked for a certain flight actually show up. If 7 passengers are randomly selected, find the probability that at least 5 of them show up.

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

(If necessary, consult a list of formulas.)

To solve the problem, we can use the binomial probability formula. The problem deals with binomial distribution because there are two possible outcomes for each passenger: either they show up or they don't.

### Step 1: Define the parameters
- \( n = 7 \): The number of passengers selected.
- \( p = 0.7 \): The probability that a passenger shows up.
- \( q = 1 - p = 0.3 \): The probability that a passenger does not show up.

We are asked to find the probability that at least 5 passengers show up. This is equivalent to finding the probability that \( X \geq 5 \), where \( X \) is the number of passengers who show up.

### Step 2: Use the binomial probability formula
The binomial probability formula is given by:

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

Where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
- \( k \) is the number of passengers that show up.

We need to calculate the probabilities for \( X = 5 \), \( X = 6 \), and \( X = 7 \), and then sum them up.

\[
P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)
\]

### Step 3: Calculate the individual probabilities

**For \( X = 5 \):**
\[
P(X = 5) = \binom{7}{5} (0.7)^5 (0.3)^{2}
\]
\[
P(X = 5) = \frac{7!}{5!(7-5)!} (0.7)^5 (0.3)^2 = \frac{7 \times 6}{2 \times 1} \times 0.16807 \times 0.09
\]
\[
P(X = 5) \approx 21 \times 0.16807 \times 0.09 \approx 0.3172
\]

**For \( X = 6 \):**
\[
P(X = 6) = \binom{7}{6} (0.7)^6 (0.3)^{1}
\]
\[
P(X = 6) = \frac{7!}{6!(7-6)!} (0.7)^6 (0.3)^1 = 7 \times 0.117649 \times 0.3
\]
\[
P(X = 6) \approx 7 \times 0.117649 \times 0.3 \approx 0.2471
\]

**For \( X = 7 \):**
\[
P(X = 7) = \binom{7}{7} (0.7)^7 (0.3)^{0}
\]
\[
P(X = 7) = 1 \times 0.0823543 \times 1 = 0.0823543
\]
\[
P(X = 7) \approx 0.0824
\]

### Step 4: Add the probabilities
\[
P(X \geq 5) = 0.3172 + 0.2471 + 0.0824 = 0.6467
\]

### Final Answer
The probability that at least 5 passengers show up is approximately **0.65** when rounded to two decimal places.

Would you like more details or have any questions? Here are some related questions to consider:

1. What is the probability that exactly 5 passengers show up?
2. What is the probability that fewer than 5 passengers show up?
3. How would the probability change if the number of passengers selected was 10?
4. How does changing the probability of a passenger showing up to 0.8 affect the result?
5. Can you determine the expected number of passengers that will show up?

**Tip:** When dealing with binomial distribution, remember that it’s ideal for situations where you have a fixed number of trials, only two possible outcomes, and a constant probability of success.

Learn how to calculate the probability that at least 5 out of 7 randomly selected passengers show up on a flight with a 70% show-up rate. Use the binomial probability formula to solve this problem step-by-step.

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