Question 1205465
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There are 14 seats. Ten of which are locked in for those guaranteed to show up for the flight.
There are 14-10 = 4 seats left and 16-10 = 6 extra people competing for those remaining seats.
These 6 extra people may or may not show up on time. 


n = 6 = sample size
p = 0.47 = probability a person arrives on time
x = number extra people who arrive on time
x is equal to a value from the set {0,1,2,3,4,5,6}
x = 2 for instance means 2 extra people arrived on time.


Random variable X follows a binonimal distribution because of these 3 reasons:<ul><li>There are two outcomes: the person shows up on time or they don't. The 2 outcomes is what the "bi" in "binomial" refers to.</li><li>Each person has the same probability of showing up on time (47% chance).</li><li>Each person's arrival is independent of one another. This assumes no one person hinders or helps any other potential passenger.</li></ul>Always check these 3 criteria to make sure we are dealing with a binomial probability distribution.


B(x) = binomial probability
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
The nCx refers to the nCr combination formula. Such values can be found in Pascal's Triangle.


Plug in the values mentioned for n and p
We'll go from this
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
to this
B(x) = (6Cx)*(0.47^x)*(0.53)^(6-x)


Then let's plug in x = 0
B(x) = (6Cx)*(0.47^x)*(0.53)^(6-x)
B(0) = (6C0)*(0.47^0)*(0.53)^(6-0)
B(0) = (1)*(0.47^0)*(0.53)^(6-0)
B(0) = 0.022164361129
B(0) = 0.0222
This is the approximate probability of having 0 extra people show up.


You could repeat similar steps for x = 1 through x = 6.


A quicker way is to use a spreadsheet. Specifically you'll use the <a href="https://support.microsoft.com/en-us/office/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c">binomDist</a> function. 
Then use the <a href="https://support.microsoft.com/en-us/office/round-function-c018c5d8-40fb-4053-90b1-b3e7f61a213c">round</a> function to do as you'd expect.
Example:  
<font color=green>=Round(A1,4)</font> will round the value at cell A1 to 4 decimal places. Don't forget about the equal sign up front. 


Here's the binomial probability distribution
<table border = "1" cellpadding = "5"><tr><td>x</td><td>B(x)</td></tr><tr><td>0</td><td>0.0222</td></tr><tr><td>1</td><td>0.1179</td></tr><tr><td>2</td><td>0.2615</td></tr><tr><td>3</td><td>0.3091</td></tr><tr><td>4</td><td>0.2056</td></tr><tr><td>5</td><td>0.0729</td></tr><tr><td>6</td><td>0.0108</td></tr></table>


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Overbooking happens when x = 5 or x = 6 extra people show up on time.
B(5) + B(6) = 0.0729+0.0108 
B(5) + B(6) = <font color=red>0.0837</font> is the approximate probability of overbooking. 
This is roughly a 8.37% chance.


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Empty seats happen when x = 0 through x = 3, i.e when x < 4.
B(0)+B(1)+B(2)+B(3) = 0.0222+0.1179+0.2615+0.3091 
B(0)+B(1)+B(2)+B(3) = <font color=red>0.7107</font>
There's roughly a 71.07% chance of the flight having one or more empty seats.


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Answers:


Probability that overbooking occurs:  <font color=red size=4>0.0837</font>
Probability that the flight has empty seats: <font color=red size=4>0.7107</font>
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