From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 9 passengers are randomly selected, find the probability that fewer than 8 of them show up.

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

To solve this problem, we need to calculate the probability that fewer than 8 out of 9 passengers show up. This is a binomial probability problem, where we can use the binomial distribution formula.

Given:
- The probability that a passenger shows up, \( p = 0.8 \).
- The probability that a passenger does not show up, \( q = 1 - p = 0.2 \).
- The number of trials (passengers), \( n = 9 \).
- We want to find the probability that fewer than 8 passengers show up, which is \( X < 8 \).

This involves calculating the probabilities for \( X = 0, 1, 2, \ldots, 7 \), and summing these probabilities.

The probability mass function for a binomial distribution is given by:

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

where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

We need to calculate:

\[
P(X < 8) = P(X = 0) + P(X = 1) + \dots + P(X = 7)
\]

Let's calculate each term:

1. **For \( X = 0 \):**
\[
P(X = 0) = \binom{9}{0} \cdot (0.8)^0 \cdot (0.2)^9 = 1 \cdot 1 \cdot 0.000000512 = 0.000000512
\]

2. **For \( X = 1 \):**
\[
P(X = 1) = \binom{9}{1} \cdot (0.8)^1 \cdot (0.2)^8 = 9 \cdot 0.8 \cdot 0.00000128 = 0.000009216
\]

3. **For \( X = 2 \):**
\[
P(X = 2) = \binom{9}{2} \cdot (0.8)^2 \cdot (0.2)^7 = 36 \cdot 0.64 \cdot 0.0000064 = 0.000147456
\]

4. **For \( X = 3 \):**
\[
P(X = 3) = \binom{9}{3} \cdot (0.8)^3 \cdot (0.2)^6 = 84 \cdot 0.512 \cdot 0.000064 = 0.002748672
\]

5. **For \( X = 4 \):**
\[
P(X = 4) = \binom{9}{4} \cdot (0.8)^4 \cdot (0.2)^5 = 126 \cdot 0.4096 \cdot 0.00032 = 0.016424448
\]

6. **For \( X = 5 \):**
\[
P(X = 5) = \binom{9}{5} \cdot (0.8)^5 \cdot (0.2)^4 = 126 \cdot 0.32768 \cdot 0.001024 = 0.042467328
\]

7. **For \( X = 6 \):**
\[
P(X = 6) = \binom{9}{6} \cdot (0.8)^6 \cdot (0.2)^3 = 84 \cdot 0.262144 \cdot 0.008 = 0.176160768
\]

8. **For \( X = 7 \):**
\[
P(X = 7) = \binom{9}{7} \cdot (0.8)^7 \cdot (0.2)^2 = 36 \cdot 0.2097152 \cdot 0.04 = 0.301989888
\]

Now, summing these probabilities:

\[
P(X < 8) = 0.000000512 + 0.000009216 + 0.000147456 + 0.002748672 + 0.016424448 + 0.042467328 + 0.176160768 + 0.301989888 = 0.539948288
\]

Rounding to two decimal places:

\[
P(X < 8) \approx 0.54
\]

So, the probability that fewer than 8 passengers show up is **0.54**.

Would you like any more details on this problem or have any questions?

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**Relative Questions:**
1. How is the binomial coefficient calculated?
2. What is the significance of the binomial distribution in probability?
3. How can we approximate the binomial distribution using a normal distribution?
4. What if the probability of a passenger showing up was different? How would that affect the calculation?
5. How does increasing the number of passengers affect the probability?
6. What if we were to calculate the probability of exactly 8 passengers showing up?
7. How does the cumulative distribution function (CDF) relate to this problem?
8. Can you explain why \( X < 8 \) was used instead of \( X \leq 7 \)?

**Tip:** Always ensure to round intermediate steps to a sufficient number of decimal places to maintain accuracy in your final answer.

Calculate the probability that fewer than 8 out of 9 passengers show up for a flight, based on known show-up rates. This problem involves using the binomial distribution to model the likelihood of passenger attendance.

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