Question 1204038: small regional carrier accepted 17 reservations for a particular flight with 13 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 50% chance, independently of each other.
(Report answers accurate to 4 decimal places.)
Probability that overbooking occurs:
Probability that the flight has empty seats
Answer by math_tutor2020(3816)   (Show Source):
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Answers:
Probability that overbooking occurs: 0.6563
Probability that the flight has empty seats: 0.1094
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Work Shown
17 reservations.
11 are locked in and will show up.
17-11 = 6 people are competing for the last 13-11 = 2 seats.
These 6 people may or may not show up.
X = number of extra people who may or may not show up for the flight.
Overbooking occurs when X is selected from the set {3,4,5,6}.
X follows a binomial probability distribution because of these reasons- There are two options: the person shows up on time or they don't.
- Each person has the same probability of showing up (50% chance)
- Each person is independent of one another.
X takes on values from the set {0,1,2,3,4,5,6}
Use the binomial probability distribution formula
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
to compute each probability.
The nCx refers to the nCr combination formula.
n = 6 people remain
p = 0.50 = probability of a person showing up
x = values chosen from the set {0,1,2,3,4,5,6}
I'll show the steps for computing B(0)
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
B(x) = (6Cx)*(0.5^x)*(1-0.5)^(6-x)
B(0) = (6C0)*(0.5^0)*(1-0.5)^(6-0)
B(0) = (1)*(0.5^0)*(1-0.5)^(6-0)
B(0) = 0.015625
This is the probability of having those last two seats empty (since 0 extra people show up).
Similar steps will be followed for B(1), B(2), all the way up to B(6)
Here's the probability distribution table.
| x | B(x) | | 0 | 0.015625 | | 1 | 0.09375 | | 2 | 0.234375 | | 3 | 0.3125 | | 4 | 0.234375 | | 5 | 0.09375 | | 6 | 0.015625 |
A spreadsheet is recommended to generate it quickly.
Various online binomial distribution calculators can be used as an alternative.
As mentioned earlier, overbooking happens when X = 3 through X = 6
B(3)+B(4)+B(5)+B(6) = 0.3125+0.234375+0.09375+0.015625 = 0.65625
Rounding to four decimal places gets us 0.6563
There's about a 65.63% chance of overbooking.
The probability of the flight having 1 or more empty seats is when X = 0 or X = 1.
B(0)+B(1) = 0.015625+0.09375 = 0.109375
That rounds to 0.1094
There's about a 10.94% chance of at least one empty seat.
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If you want to use a TI83 or TI84 calculator, then check out this article
https://www.statology.org/binomial-probabilities-ti-84-calculator/
For the first part you'll type in 1-binomCDF(6,0.5,2)
The binomCDF(6,0.5,2) portion computes B(0)+B(1)+B(2). Subtract that sum from 1 to get the result of B(3)+B(4)+B(5)+B(6).
The second part will have you type in binomCDF(6,0.5,1) to compute B(0)+B(1)
We won't be subtracting from 1 like last time.
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