Question 1204520
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Suppose that the probability that a passenger will miss a flight is 0.0999. 
Airlines do not like flights with empty​ seats, but it is also not desirable 
to have overbooked flights because passengers must be​ "bumped" from the flight. 
Suppose that an airplane has a seating capacity of 57 passengers.
​(a) If 59 tickets are​ sold, what is the probability that 58 or 59 passengers show up 
for the flight resulting in an overbooked​ flight?
​(b) Suppose that 63 tickets are sold. What is the probability that a passenger 
will have to be​ "bumped"?
​(c) For a plane with seating capacity of 250 ​passengers, what is the largest number of tickets 
that can be sold to keep the probability of a passenger being​ "bumped" below 1​%?
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<pre>

                 <U>Question  (a)</U>


It is a binomial experiment.


There are 59 potential passengers. The probability that every given passenger will show up is 1 - 0.0999 = 0.9001.

The events for individual passengers are INDEPENDENT.

The question is: find the probability that of 59 passengers that bought tickets, 58 or 59 will show up
(it is the event "the flight is overbooked").


n = 59 (the number of trials);

p = 0.9001  (the probability of "success" to each individual trial);

k >= 58  (the event of overbooking flight).


P(n=59; p=0.9001; k>=58) = P(n=59; p=0.9001; k=58) + P(n=59; p=0.9001; k=59) = {{{C[59]^58*0.9001^58*0.0999}}} + {{{C[59]^59*0.9001^59}}} =

   = {{{59*0.9001^58*0.0999 + 0.0999^59}}} = 0.01316  (rounded).    <U>ANSWER</U>



                 <U>Question  (b)</U>


It is a binomial experiment very similar to part (a).


There are 63 potential passengers. The probability that every given passenger will show up is 1 - 0.0999 = 0.9001.

The events for individual passengers are INDEPENDENT.

The question is: find the probability that of 63 passengers that bought tickets, at least 58 will show up
(it is the event "the flight is overbooked").


n = 63 (the number of trials);

p = 0.9001  (the probability of "success" to each individual trial);

k >= 58  (the event of overbooking flight).


We want to find the probability  P = P(n=63; p=0.9001; k>=58) 


Go to web-site https://stattrek.com/online-calculator/binomial and use online (free of charge) calculator there.
Its interface is very simple, so even a beginner student can easily work with it.
It facilitates using  many-addend formula, so calculations are really easy and fast.


The <U>ANSWER</U> is

       P = P(n=63; p=0.9001; k>=58)  = 0.3893.   


Alternatively, you may use standard function binomcfd on your regular TI-83/84 calculator.

About this function and how to use it read from this web-page https://www.statology.org/binompdf-vs-binomcdf/

The regular calculator will provide the same answer.

</pre>

Solved.


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My post is just very long, so I prefer to stop here.