This problem involves calculating the probability of fewer than a certain number of passengers showing up for a flight, given a specific probability of each passenger showing up. The situation can be modeled using the binomial distribution, as the problem involves a fixed number of trials (9 passengers), with only two possible outcomes for each trial (showing up or not).

Given:

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

Step 1: Identify the scenario.

The problem asks for the probability that fewer than $k = 5$ passengers show up.

Step 2: Apply 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:

• $X$ is the number of passengers who show up.
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.
• $p$ is the probability of success (a passenger showing up).
• $1-p$ is the probability of failure (a passenger not showing up).

Step 3: Calculate the cumulative probability.

We need to find $P(X < 5)$, which means we sum the probabilities for $X = 0$ to $X = 4$:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

I'll calculate these probabilities individually and sum them.

Step 4: Compute each probability.

Let's compute the values:

1. $P(X = 0)$: $P(X = 0) = \binom{9}{0} \cdot (0.80)^0 \cdot (0.20)^9 = 1 \times 1 \times 0.00000256 = 0.00000256$

2. $P(X = 1)$: $P(X = 1) = \binom{9}{1} \cdot (0.80)^1 \cdot (0.20)^8 = 9 \times 0.8 \times 0.000065536 = 0.00047186$

3. $P(X = 2)$: $P(X = 2) = \binom{9}{2} \cdot (0.80)^2 \cdot (0.20)^7 = 36 \times 0.64 \times 0.000524288 = 0.01210368$

4. $P(X = 3)$: $P(X = 3) = \binom{9}{3} \cdot (0.80)^3 \cdot (0.20)^6 = 84 \times 0.512 \times 0.0016777216 = 0.07103232$

5. $P(X = 4)$: $P(X = 4) = \binom{9}{4} \cdot (0.80)^4 \cdot (0.20)^5 = 126 \times 0.4096 \times 0.01048576 = 0.5407314$

Step 5: Sum the probabilities.

$P(X < 5) = 0.00000256 + 0.00047186 + 0.01210368 + 0.07103232 + 0.5407314 = 0.62434182$

Step 6: Round the final answer to four decimal places.

The probability that fewer than 5 passengers show up is approximately:

$\boxed{0.6243}$

Do you want further details or have any questions?

Relative Questions:

1. How would the probability change if the number of selected passengers increased to 10?
2. What would be the probability that exactly 6 passengers show up?
3. How is the binomial distribution different from a normal distribution?
4. Can this problem be solved using a normal approximation? If so, how?
5. What if the probability of showing up was 90% instead of 80%?
6. How would the scenario change if we were interested in 5 or more passengers showing up?
7. What role does the complement rule play in calculating these types of probabilities?
8. Can you calculate the expected number of passengers who will show up for this flight?

Tip:

When dealing with binomial probabilities, remember that cumulative probability calculations (like $P(X \leq k)$) can often be simplified using statistical software or binomial tables.