SOLUTION: Suppose that the probability that a passenger will miss a flight is 0.0957. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flight
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Question 1161858: Suppose that the probability that a passenger will miss a flight is 0.0957. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 59 passengers.
(a) If 61 tickets are sold, what is the probability that 60 or 61 passengers show up for the flight resulting in an overbooked flight?
(b) Suppose that 65 tickets are sold. What is the probability that a passenger will have to be "bumped"?
(c) For a plane with seating capacity of
52 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 5%?
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
from scipy.stats import binom
# Constants
p_miss = 0.0957
p_show = 1 - p_miss
# (a) n=61, capacity=59, find P(k=60) + P(k=61)
n_a = 61
p_60 = binom.pmf(60, n_a, p_show)
p_61 = binom.pmf(61, n_a, p_show)
ans_a = p_60 + p_61
# (b) n=65, capacity=59, find P(k > 59)
n_b = 65
capacity_b = 59
ans_b = binom.sf(capacity_b, n_b, p_show) # sf is 1 - cdf, so P(X > 59)
# (c) capacity=52, find max n such that P(k > 52) < 0.05
capacity_c = 52
threshold = 0.05
best_n = 52
for n in range(53, 100):
prob_bumped = binom.sf(capacity_c, n, p_show)
if prob_bumped < threshold:
best_n = n
else:
break
print(f"Part (a): {ans_a=}")
print(f"Part (b): {ans_b=}")
print(f"Part (c): {best_n=}, {binom.sf(capacity_c, best_n, p_show)=}, {binom.sf(capacity_c, best_n+1, p_show)=}")
```
```text?code_stdout&code_event_index=2
Part (a): ans_a=0.016126517739814418
Part (b): ans_b=0.4014073805745022
Part (c): best_n=54, binom.sf(capacity_c, best_n, p_show)=0.029369939359098, binom.sf(capacity_c, best_n+1, p_show)=0.09276094534760537
```
To solve these problems, we use the **Binomial Distribution**.
* Let $n$ be the number of tickets sold.
* Let $p$ be the probability that a passenger shows up for the flight. Given the probability of missing is $0.0957$, the probability of showing up is $p = 1 - 0.0957 = 0.9043$.
* Let $X$ be the number of passengers who show up. $X \sim \text{Binomial}(n, 0.9043)$.
---
### (a) If 61 tickets are sold, what is the probability that 60 or 61 passengers show up?
We need to calculate $P(X=60) + P(X=61)$ for $n=61$.
Using the binomial formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$:
* $P(X=60) = \binom{61}{60} (0.9043)^{60} (0.0957)^1 \approx 0.0139$
* $P(X=61) = \binom{61}{61} (0.9043)^{61} (0.0957)^0 \approx 0.0022$
$$P(60 \text{ or } 61) = 0.01392 + 0.00221 \approx 0.0161$$
**Probability: 0.0161**
---
### (b) If 65 tickets are sold, what is the probability that a passenger will be "bumped"?
A passenger is bumped if more than 59 people show up ($X > 59$). For $n=65$:
$$P(X > 59) = P(X=60) + P(X=61) + P(X=62) + P(X=63) + P(X=64) + P(X=65)$$
Using the cumulative distribution function (or summing the individual probabilities):
$$P(X > 59) \approx 0.4014$$
**Probability: 0.4014**
---
### (c) For a capacity of 52, how many tickets may be sold to keep the probability of being "bumped" below 5%?
We need to find the largest $n$ such that $P(X > 52) < 0.05$.
* If **$n=53$** tickets are sold: $P(X > 52) = P(X=53) = (0.9043)^{53} \approx 0.0050$. (This is below 5%)
* If **$n=54$** tickets are sold: $P(X > 52) = P(X=53) + P(X=54) \approx 0.0294$. (This is below 5%)
* If **$n=55$** tickets are sold: $P(X > 52) \approx 0.0928$. (This exceeds 5%)
**Maximum tickets that can be sold: 54**
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