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 fewer than 6 of them show up.

To solve this problem, we'll model it using the binomial distribution, which is appropriate when there are a fixed number of trials (passengers) and two possible outcomes for each trial (showing up or not showing up).

### Step 1: Define the binomial distribution 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 means we need to calculate the probability for \(X < 6\).

### Step 2: Calculate the individual probabilities
For \(X\) to be the number of passengers that show up, \(X\) follows a binomial distribution:

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

We'll calculate the probabilities for \(X = 0, 1, 2, 3, 4, 5\).

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:

\[
P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

\[
P(X < 6) = 0.000128 + 0.003584 + 0.043008 + 0.183501 + 0.3670016 + 0.2752512 = 0.8724738
\]

So, the probability that fewer than 6 passengers show up is approximately **0.872**.

Would you like to see more details on how to compute binomial probabilities or do you have any other questions?

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Here are some related questions:
1. What is the probability that exactly 6 passengers show up?
2. What is the expected number of passengers who show up for this flight?
3. How does the probability change if the number of passengers is increased to 10?
4. What is the variance of the number of passengers showing up?
5. How would the probability change if the show-up rate increased to 90%?
6. What is the probability that all 7 passengers show up?
7. How does this problem relate to the law of large numbers?
8. Can we use the normal approximation for this binomial distribution?

**Tip:** Understanding binomial distribution is crucial in scenarios where there are a fixed number of independent trials, each with two possible outcomes, like this passenger show-up problem.

Explore how to calculate the probability that fewer than 6 out of 7 randomly selected passengers show up for a flight, given an 80% show-up rate. Learn about binomial distribution, key formulas, and step-by-step solutions.

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