Question 1204038
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Answers:
Probability that overbooking occurs: <font color=red size=4>0.6563</font>
Probability that the flight has empty seats: <font color=red size=4>0.1094</font>


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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<ul><li>There are two options: the person shows up on time or they don't.</li><li>Each person has the same probability of showing up (50% chance)</li><li>Each person is independent of one another.</li></ul>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.
<table border = "1" cellpadding = "5"><tr><td>x</td><td>B(x)</td></tr><tr><td>0</td><td>0.015625</td></tr><tr><td>1</td><td>0.09375</td></tr><tr><td>2</td><td>0.234375</td></tr><tr><td>3</td><td>0.3125</td></tr><tr><td>4</td><td>0.234375</td></tr><tr><td>5</td><td>0.09375</td></tr><tr><td>6</td><td>0.015625</td></tr></table>
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 <font color=red>0.6563</font>
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 <font color=red>0.1094</font>
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
<a href = "https://www.statology.org/binomial-probabilities-ti-84-calculator/">https://www.statology.org/binomial-probabilities-ti-84-calculator/</a>


For the first part you'll type in <font color=red>1-binomCDF(6,0.5,2)</font>
The <font color=red>binomCDF(6,0.5,2)</font> 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 <font color=red>binomCDF(6,0.5,1)</font> to compute B(0)+B(1)
We won't be subtracting from 1 like last time. 
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