Airlines often sell more tickets for a flight than there are seats because some ticket holders do not show up for the flight. Assume that an airplane has 170 seats for passengers and that the probability that a person holding a ticket appears for the flight is 0.93. If the airline sells 176 tickets, what is the probability that everyone who appears for the flight will get a seat?

Guest Jul 3, 2014

#2**+8 **

$${\mathtt{1}}{\mathtt{\,-\,}}\left({\left({\frac{{\mathtt{176}}{!}}{{\mathtt{171}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{171}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{171}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{172}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{172}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{172}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{173}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{173}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{173}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{174}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{174}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{174}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{175}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{175}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{175}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.93}}}^{{\mathtt{176}}}\right) = {\mathtt{0.986\: \!001\: \!218\: \!893\: \!262\: \!3}}$$

$$P(enough\: seats )\approx 0.9860$$

I am reasonably sure that mine is correct.

Melody
Jul 3, 2014

#2**+8 **

Best Answer

$${\mathtt{1}}{\mathtt{\,-\,}}\left({\left({\frac{{\mathtt{176}}{!}}{{\mathtt{171}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{171}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{171}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{5}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{172}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{172}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{172}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{173}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{173}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{173}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{174}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{174}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{174}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{176}}{!}}{{\mathtt{175}}{!}{\mathtt{\,\times\,}}({\mathtt{176}}{\mathtt{\,-\,}}{\mathtt{175}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.93}}}^{{\mathtt{175}}}{\mathtt{\,\times\,}}{{\mathtt{0.07}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.93}}}^{{\mathtt{176}}}\right) = {\mathtt{0.986\: \!001\: \!218\: \!893\: \!262\: \!3}}$$

$$P(enough\: seats )\approx 0.9860$$

I am reasonably sure that mine is correct.

Melody
Jul 3, 2014

#4**+5 **

This is an amazing piece of work, Melody. You typed this out like someone playing Vanderbeck’s "Edelweiss Glide" waltz on the piano. I would have just popped in a CD.

With modern computer technology, it is easy to forget the complicated mathematics behind the functions, and just type in three or four numbers and accept the answer the computer generates without giving a second thought to the process behind it.

Depending on the computer program, it may look this way:

P(x <= 170) = binomcdf(176,0.93,170)= 0.98600121889…

This is great. … But …

This causes a loss of the intuitive sense of the process that develops when solving the equations manually.

Here is an overview of Melody’s output, for any who may be interested.

The basic formulas are twofold:

First, the binomial distribution:

This formula generates the discrete probability distribution of successes (or failures) in a sequence of n binary (0 or 1) independent trials. This yields success with probability p.

The general binomial formula:

$$\displaystyle \Pr \left({X = k}\right) = \binom n k p^k \left({1-p}\right)^{n-k}$$

This defines k successes (p^{k}) and (n − k) failures (1 − p)^{(n–k)} multiplied by a binomial coefficient yielding a probability value as its product.

The successes (or failures) can occur anywhere among the n trials. The n trials are **not** linear, they are hypergeometric and the sample space can change in very large, non-linear, values with small change in population. The binomial component distributes this over the sample space, and effectively normalizes this to a monotone event and generates a density or mass function, with a point of central tendency.

Second, the CDF – Cumulative Distribution Function:

This question requires a CDF. CDFs can have complicated theoretical applications (Translation: I do not fully understand it; so, how can I explain it). However, in this case, this is straightforward; simply add the individual terms. This totals all the individual functions and sets a boundary. In this case, the boundary is subtracted from one to give the probability of the sample group having a seat, instead of not having a seat.

$$\:\\ \ \hspace{150pt} \ \Pr \left({X =< 170}\right)=\\

\ 1- \

\displaystyle \binom {176} {171} {0.93}^{171} \left({0.07}\right)^{5}+

\displaystyle \binom {176} {172} {0.93}^{172} \left({0.07}\right)^{4}+

\displaystyle \binom {176} {173} {0.93}^{173} \left({0.07}\right)^{3}\\

\displaystyle \ +\ \binom {176} {174} {0.93}^{174} \left({0.07}\right)^{2}+

\displaystyle \binom {176} {175} {0.93}^{175} \left({0.07}\right)^{1}+

\displaystyle \binom {176} {176} {0.93}^{176} \left({0.07}\right)^{0}\\\

\:\\ \ \hspace{150pt} \ = 0.98600121889 \\$$

(A reasonable number of decimal places are included here).

Note the last binomial and exponent of 0 resolve to a value of 1. They are included for continuity. With the above data, other observations are statically discernable.

-----

Here is an example of where a discreet binomial is used (this does not require a CDF):

What is the probability that the flight departs with exactly 5 empty seats?

$$\displaystyle \binom {176} {166} {0.93}^{166} \left({0.07}\right)^{10}

\hspace{5pt} \ = \ 0.100033 \\$$

Often, in this forum, you are referred to in terms of royalty: a queen. Here is an honorary title earned and bestowed with honor and respect, because of your skill, instead of the randomness of a birthright.

I concur!

~~D~~

DavidQD
Jul 4, 2014