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.