
Aeroplane Question!!
i) Eagle Airline is a small commuter airline.Their planes hold only 15 people.
Past records indicate that 30% of people making a reservation do not show
up for the flight. Suppose that Eagle Air decide to book 20 people for
each flight. Determne the probability that on any given flight,at least one
passenger holding a reservation will not have a seat?
ii) Suppose on a particular flight from Dallas to El Paso,Eagle Air uses a much
larger aircraft that holds 200 people. As in the last question,still 30% of
the people who make reservations don't show up. If the airline accepts 265
reservations,what is the probability that at least one passenger will end up
without a seat on this flight?

it's funny that you should ask this questions since I had this word for word on my stats final. Both of these are binomial distribution.
1) $\displaystyle P(X \geq 21) = P(X = 21) \cup P(X= 22) \cup P(X=23) \cup P(X=24) \cup P(X=25) $
$\displaystyle {{20}\choose{16}} (0.3)^{16}(0.7)^4 = $
$\displaystyle {{20}\choose{17}} (0.3)^{17}(0.7)^3 = $
$\displaystyle {{20}\choose{18}} (0.3)^{18}(0.7)^2 = $
$\displaystyle {{20}\choose{19}} (0.3)^{19}(0.7)^1 = $
$\displaystyle {{20}\choose{20}} (0.3)^{20}(0.7)^0 = $
2) it would basically the same principle as the first one, so you would have: $\displaystyle P(X \geq 201) = P(X = 201) \cup P(X= 202) \cup P(X=203) \cup ... \cup P(X=265)$
$\displaystyle {{265}\choose{201}} (0.3)^{201}(0.7)^{64} = $
$\displaystyle \vdots $
$\displaystyle {{265}\choose{265}} (0.3)^{265}(0.7)^{0} = $
you could set up excel for the second one since doing it by hand would take awhile