To solve this problem with higher precision, we will recalculate the probabilities with at least four decimal places and round the final answer to two decimal places.

Step 1: Define the parameters

• n: Number of trials (passengers) = 7
• p: Probability of success (showing up) = 0.80
• q: Probability of failure (not showing up) = 1 - p = 0.20

We want to find the probability that fewer than 6 passengers show up, which corresponds to $P(X < 6)$.

Step 2: Calculate individual probabilities

Using the binomial probability formula: $P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}$ where $\binom{n}{k}$ is the binomial coefficient.

1. $P(X = 0)$: $P(X = 0) = \binom{7}{0} \cdot (0.80)^0 \cdot (0.20)^7 = 1 \cdot 1 \cdot 0.000128 = 0.000128$

2. $P(X = 1)$: $P(X = 1) = \binom{7}{1} \cdot (0.80)^1 \cdot (0.20)^6 = 7 \cdot 0.80 \cdot 0.00064 = 0.003584$

3. $P(X = 2)$: $P(X = 2) = \binom{7}{2} \cdot (0.80)^2 \cdot (0.20)^5 = 21 \cdot 0.64 \cdot 0.0032 = 0.043008$

4. $P(X = 3)$: $P(X = 3) = \binom{7}{3} \cdot (0.80)^3 \cdot (0.20)^4 = 35 \cdot 0.512 \cdot 0.01024 = 0.183501$

5. $P(X = 4)$: $P(X = 4) = \binom{7}{4} \cdot (0.80)^4 \cdot (0.20)^3 = 35 \cdot 0.4096 \cdot 0.0256 = 0.3670016$

6. $P(X = 5)$: $P(X = 5) = \binom{7}{5} \cdot (0.80)^5 \cdot (0.20)^2 = 21 \cdot 0.32768 \cdot 0.04 = 0.2752512$

Step 3: Sum the probabilities for $X < 6$

The probability that fewer than 6 passengers show up is: $P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$ $P(X < 6) = 0.000128 + 0.003584 + 0.043008 + 0.183501 + 0.3670016 + 0.2752512 = 0.8724738$

Step 4: Round the final answer

Rounding 0.8724738 to two decimal places gives:

$P(X < 6) \approx 0.87$

So, the probability that fewer than 6 passengers show up is 0.87.

Would you like any more details or further assistance with this problem?

---

Here are some related questions:

1. What is the probability that exactly 6 passengers show up?
2. What is the probability that all 7 passengers show up?
3. How does the probability change if the probability of showing up is reduced to 70%?
4. What is the expected number of passengers who will show up?
5. What is the variance and standard deviation of the number of passengers who show up?
6. How would the results change if the number of passengers was increased to 10?
7. What is the probability that at least 5 passengers show up?
8. How does this binomial distribution relate to the normal distribution for larger sample sizes?

Tip: When dealing with probabilities, maintaining precision in intermediate steps helps ensure accuracy in the final result, especially when rounding.