From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 7 passengers are randomly selected, find the probability that at most 4 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'll use the binomial probability formula since the situation involves a fixed number of independent trials, each with two possible outcomes (either a passenger shows up or doesn't show up). The binomial probability formula is:

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

Where:
- \( n \) is the number of trials (in this case, 7 passengers).
- \( k \) is the number of successes (in this case, the number of passengers who show up).
- \( p \) is the probability of a success on an individual trial (in this case, 0.80, since 80% of the passengers show up).
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

### Step 1: Calculate the probability for each scenario where at most 4 passengers show up.

We need to find the probability that at most 4 passengers show up, so we calculate the sum of probabilities for \( k = 0, 1, 2, 3, 4 \).

\[
P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\]

### Step 2: Compute the individual probabilities.

1. **\( P(X = 0) \):**
   \[
   P(X = 0) = \binom{7}{0} \cdot 0.8^0 \cdot 0.2^7 = 1 \cdot 1 \cdot 0.000128 = 0.000128
   \]

2. **\( P(X = 1) \):**
   \[
   P(X = 1) = \binom{7}{1} \cdot 0.8^1 \cdot 0.2^6 = 7 \cdot 0.8 \cdot 0.00064 = 0.003584
   \]

3. **\( P(X = 2) \):**
   \[
   P(X = 2) = \binom{7}{2} \cdot 0.8^2 \cdot 0.2^5 = 21 \cdot 0.64 \cdot 0.0032 = 0.043008
   \]

4. **\( P(X = 3) \):**
   \[
   P(X = 3) = \binom{7}{3} \cdot 0.8^3 \cdot 0.2^4 = 35 \cdot 0.512 \cdot 0.01024 = 0.1835008
   \]

5. **\( P(X = 4) \):**
   \[
   P(X = 4) = \binom{7}{4} \cdot 0.8^4 \cdot 0.2^3 = 35 \cdot 0.4096 \cdot 0.008 = 0.114688
   \]

### Step 3: Sum the probabilities.

\[
P(X \leq 4) = 0.000128 + 0.003584 + 0.043008 + 0.1835008 + 0.114688 = 0.3449088
\]

### Step 4: Round the result to two decimal places.

\[
P(X \leq 4) \approx 0.34
\]

Thus, the probability that at most 4 passengers out of 7 will show up is **0.34**.

Would you like any further details or have any questions?

---

Here are some related questions:
1. How do we interpret the binomial coefficient \(\binom{n}{k}\)?
2. What would be the probability if exactly 5 passengers show up?
3. How would the probability change if the show-up rate was 90% instead of 80%?
4. How is the mean and variance of a binomial distribution calculated?
5. Can this binomial distribution be approximated by a normal distribution?
6. What is the cumulative probability that at least 3 passengers show up?
7. How would the probability change if the number of passengers selected was increased to 10?
8. What is the effect of increasing the number of trials \(n\) on the binomial distribution?

**Tip:** When calculating binomial probabilities, it is helpful to understand the concept of cumulative probabilities, especially for calculating "at most" or "at least" probabilities.

Learn how to calculate the probability that at most 4 out of 7 passengers show up for a flight, using the binomial probability formula and combinatorial mathematics. Get detailed step-by-step solutions with intermediate computations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation