document.write( "Question 1205465: User
\n" ); document.write( "A small regional carrier accepted 16 reservations for a particular flight with 14 seats. 10 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 47% chance, independently of each other.
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\n" ); document.write( "\n" ); document.write( "Find the probability that overbooking occurs.
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\n" ); document.write( "Find the probability that the flight has empty seats\r
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Algebra.Com's Answer #842288 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "A similar question has been asked before.
\n" ); document.write( "Check out this link
\n" ); document.write( "https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1204038.html\r
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\n" ); document.write( "\n" ); document.write( "If you're still stuck, then read on.\r
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\n" ); document.write( "\n" ); document.write( "There are 14 seats. Ten of which are locked in for those guaranteed to show up for the flight.
\n" ); document.write( "There are 14-10 = 4 seats left and 16-10 = 6 extra people competing for those remaining seats.
\n" ); document.write( "These 6 extra people may or may not show up on time. \r
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\n" ); document.write( "\n" ); document.write( "n = 6 = sample size
\n" ); document.write( "p = 0.47 = probability a person arrives on time
\n" ); document.write( "x = number extra people who arrive on time
\n" ); document.write( "x is equal to a value from the set {0,1,2,3,4,5,6}
\n" ); document.write( "x = 2 for instance means 2 extra people arrived on time.\r
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\n" ); document.write( "\n" ); document.write( "Random variable X follows a binonimal distribution because of these 3 reasons:
  • 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.
  • Each person has the same probability of showing up on time (47% chance).
  • Each person's arrival is independent of one another. This assumes no one person hinders or helps any other potential passenger.
Always check these 3 criteria to make sure we are dealing with a binomial probability distribution.\r
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\n" ); document.write( "\n" ); document.write( "B(x) = binomial probability
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "The nCx refers to the nCr combination formula. Such values can be found in Pascal's Triangle.\r
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\n" ); document.write( "\n" ); document.write( "Plug in the values mentioned for n and p
\n" ); document.write( "We'll go from this
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "to this
\n" ); document.write( "B(x) = (6Cx)*(0.47^x)*(0.53)^(6-x)\r
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\n" ); document.write( "\n" ); document.write( "Then let's plug in x = 0
\n" ); document.write( "B(x) = (6Cx)*(0.47^x)*(0.53)^(6-x)
\n" ); document.write( "B(0) = (6C0)*(0.47^0)*(0.53)^(6-0)
\n" ); document.write( "B(0) = (1)*(0.47^0)*(0.53)^(6-0)
\n" ); document.write( "B(0) = 0.022164361129
\n" ); document.write( "B(0) = 0.0222
\n" ); document.write( "This is the approximate probability of having 0 extra people show up.\r
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\n" ); document.write( "\n" ); document.write( "You could repeat similar steps for x = 1 through x = 6.\r
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\n" ); document.write( "\n" ); document.write( "A quicker way is to use a spreadsheet. Specifically you'll use the binomDist function.
\n" ); document.write( "Then use the round function to do as you'd expect.
\n" ); document.write( "Example:
\n" ); document.write( "=Round(A1,4) will round the value at cell A1 to 4 decimal places. Don't forget about the equal sign up front. \r
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\n" ); document.write( "\n" ); document.write( "Here's the binomial probability distribution
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xB(x)
00.0222
10.1179
20.2615
30.3091
40.2056
50.0729
60.0108
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\n" ); document.write( "\n" ); document.write( "Overbooking happens when x = 5 or x = 6 extra people show up on time.
\n" ); document.write( "B(5) + B(6) = 0.0729+0.0108
\n" ); document.write( "B(5) + B(6) = 0.0837 is the approximate probability of overbooking.
\n" ); document.write( "This is roughly a 8.37% chance.\r
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\n" ); document.write( "\n" ); document.write( "Empty seats happen when x = 0 through x = 3, i.e when x < 4.
\n" ); document.write( "B(0)+B(1)+B(2)+B(3) = 0.0222+0.1179+0.2615+0.3091
\n" ); document.write( "B(0)+B(1)+B(2)+B(3) = 0.7107
\n" ); document.write( "There's roughly a 71.07% chance of the flight having one or more empty seats.\r
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\n" ); document.write( "\n" ); document.write( "Answers:\r
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\n" ); document.write( "\n" ); document.write( "Probability that overbooking occurs: 0.0837
\n" ); document.write( "Probability that the flight has empty seats: 0.7107
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