Question 1205465: User
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.
(Report answers accurate to 4 decimal places.)
Find the probability that overbooking occurs.
Find the probability that the flight has empty seats
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
A similar question has been asked before.
Check out this link
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1204038.html
If you're still stuck, then read on.
----------------------------------------------------------------------
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:- 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.
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 binomDist function.
Then use the round function to do as you'd expect.
Example:
=Round(A1,4) 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
| x | B(x) | | 0 | 0.0222 | | 1 | 0.1179 | | 2 | 0.2615 | | 3 | 0.3091 | | 4 | 0.2056 | | 5 | 0.0729 | | 6 | 0.0108 |
-----------------------------------
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) = 0.0837 is the approximate probability of overbooking.
This is roughly a 8.37% chance.
-----------------------------------
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) = 0.7107
There's roughly a 71.07% chance of the flight having one or more empty seats.
----------------------------------------------------------------------
----------------------------------------------------------------------
Answers:
Probability that overbooking occurs: 0.0837
Probability that the flight has empty seats: 0.7107
Answer by ikleyn(52852)  (Show Source):
You can put this solution on YOUR website! .
Similar problems were solved at this forum before under these links
https://www.algebra.com/statistics/Binomial-probability/Binomial-probability.faq.question.1204476.html
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1204520.html
Consider them as TEMPLATES and solve this given problem in the same way.
|
|
|
