SOLUTION: 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 remaini

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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): You can put this solution on YOUR website!

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 reasonsX 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.
xB(x)
00.015625
10.09375
20.234375
30.3125
40.234375
50.09375
60.015625

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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