To find the probability that fewer than 8 out of 9 passengers show up for a flight (where each passenger has an 80% chance of showing up), we again use the binomial distribution.

Given:

• Probability of a passenger showing up, $p = 0.80$
• Number of passengers, $n = 9$
• We want to find the probability that fewer than 8 passengers show up.

Step 1: Identify the required probability.

We need to find $P(X < 8)$, which is the sum of the probabilities from $X = 0$ to $X = 7$.

Step 2: Use the binomial probability formula.

The binomial probability mass function is:

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

Where $k$ ranges from 0 to 7.

Step 3: Compute each probability.

Let’s compute the probabilities for $X = 0$ through $X = 7$ and sum them.

1. $P(X = 0)$: $P(X = 0) = \binom{9}{0} \cdot (0.80)^0 \cdot (0.20)^9 = 1 \times 1 \times 0.000000512 = 0.000000512$

2. $P(X = 1)$: $P(X = 1) = \binom{9}{1} \cdot (0.80)^1 \cdot (0.20)^8 = 9 \times 0.8 \times 0.0000032768 = 0.0000236928$

3. $P(X = 2)$: $P(X = 2) = \binom{9}{2} \cdot (0.80)^2 \cdot (0.20)^7 = 36 \times 0.64 \times 0.00002048 = 0.0004718592$

4. $P(X = 3)$: $P(X = 3) = \binom{9}{3} \cdot (0.80)^3 \cdot (0.20)^6 = 84 \times 0.512 \times 0.00008192 = 0.0035327898$

5. $P(X = 4)$: $P(X = 4) = \binom{9}{4} \cdot (0.80)^4 \cdot (0.20)^5 = 126 \times 0.4096 \times 0.00032768 = 0.0167241734$

6. $P(X = 5)$: $P(X = 5) = \binom{9}{5} \cdot (0.80)^5 \cdot (0.20)^4 = 126 \times 0.32768 \times 0.00128 = 0.0521070592$

7. $P(X = 6)$: $P(X = 6) = \binom{9}{6} \cdot (0.80)^6 \cdot (0.20)^3 = 84 \times 0.262144 \times 0.0064 = 0.1402410086$

8. $P(X = 7)$: $P(X = 7) = \binom{9}{7} \cdot (0.80)^7 \cdot (0.20)^2 = 36 \times 0.2097152 \times 0.04 = 0.301989888$

Step 4: Sum the probabilities.

$P(X < 8) = 0.000000512 + 0.0000236928 + 0.0004718592 + 0.0035327898 + 0.0167241734 + 0.0521070592 + 0.1402410086 + 0.301989888 = 0.515091$

Step 5: Round the final answer to two decimal places.

The probability that fewer than 8 passengers show up is approximately:

$\boxed{0.52}$

Would you like more details or have any other questions?

Relative Questions:

1. What would be the probability that exactly 8 passengers show up?
2. How would the probability change if the number of selected passengers increased to 10?
3. What if the probability of showing up was 90% instead of 80%?
4. How does this binomial distribution relate to a cumulative distribution function (CDF)?
5. How could you approximate this problem using a normal distribution?
6. What would the probability be if you were interested in exactly 9 passengers showing up?
7. Can this problem be extended to find the expected number of no-shows?
8. How would the calculations change if passengers were not independent in their decision to show up?

Tip:

For binomial distribution problems, using a cumulative distribution function (CDF) or statistical software can simplify the process of finding probabilities involving multiple sums, especially for larger values of $n$.