This problem involves finding the probability that at least 4 out of 6 passengers show up for a flight, given that each passenger has an 85% chance of showing up. This is a binomial probability problem.

Binomial Distribution Formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ Where:

• $n$ is the number of trials (passengers) = 6
• $k$ is the number of successes
• $p$ is the probability of success on each trial = 0.85
• $1-p$ is the probability of failure = 0.15

We need to find the probability that at least 4 passengers show up, which means we need to sum the probabilities from $P(X = 4)$ to $P(X = 6)$:

$P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)$

Step-by-Step Calculations:

1. Calculate $P(X = 4)$: $P(X = 4) = \binom{6}{4} (0.85)^4 (0.15)^2 = \frac{6!}{4!2!} \times (0.85)^4 \times (0.15)^2$ $P(X = 4) = 15 \times 0.5220 \times 0.0225 = 0.1760$

2. Calculate $P(X = 5)$: $P(X = 5) = \binom{6}{5} (0.85)^5 (0.15)^1 = \frac{6!}{5!1!} \times (0.85)^5 \times 0.15$ $P(X = 5) = 6 \times 0.4437 \times 0.15 = 0.3993$

3. Calculate $P(X = 6)$: $P(X = 6) = \binom{6}{6} (0.85)^6 (0.15)^0 = 1 \times (0.85)^6 \times 1 = 0.3771$

Sum the Probabilities:

$P(X \geq 4) = 0.1760 + 0.3993 + 0.3771 = 0.9524$

Final Answer:

The probability that at least 4 out of 6 passengers show up is approximately 0.95 when rounded to two decimal places.

Would you like more details, or do you have any questions about this problem?

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Related Questions:

1. What is the probability that exactly 4 passengers show up?
2. How does the probability change if the number of selected passengers increases to 10?
3. What is the probability that none of the 6 passengers show up?
4. What is the expected number of passengers that will show up?
5. What is the variance of the number of passengers that show up?
6. How would the probability change if the show-up rate was 90% instead of 85%?
7. Can you use a normal approximation to estimate this probability? If so, how?
8. What is the probability that fewer than 4 passengers show up?

Tip: In binomial problems, consider whether you're asked for "at least" or "at most" cases; this helps in identifying which probabilities to sum.