I'm given the formula:
$\displaystyle P(X = k; n, p) = \binom {n}{k} * p^k * q^{n-k}$
And we need to work out the binomial coefficient by hand, instead of using C(n,r).
So I have a question:
"Some airlines deliberately overbook seats, safe in
the knowledge that past experience has shown that there will
always be “no show” passengers. Let us say that an aircraft has
a capacity of 150. The airline has a policy of selling 160 tickets to
protect against no-show passengers based on past experience
that the probability of a passenger being a no show is 0.1. What
is the probability that at least one passenger will have to be
ʻbumpedʼ from the flight?"
So deducing from this I have set the following:
Let showing up = success = p = 0.9
Let not showing up = q = failure = 0.1
n = 160 and k = 150
Inputting this in to the formula I get:
$\displaystyle P(X = 150; 160, 0.9) = \binom {160}{150} * 0.9^{150} * 0.1^{10}$
To make it easier to do by hand I change the binomial coefficient:
$\displaystyle P(X = 150; 160, 0.9) = \binom {160}{10} * 0.9^{150} * 0.1^{10}$
Then:
$\displaystyle \frac{160*159*158*157*156*155*154*153*152*151}{1*2*3*4*5*6*7*8*9*10} * 0.9^{150} * 0.1^{10}$
…
$= 0.03112979$
I believe I'm using the binomial distribution correctly (or am I?) – But this is not the answer they want. They want 0.0359, and they put the probability in the form P(X > 150).
I have no idea how I can calculate this, without using equals. I thought "well, 151 is > 150, so I'll input that instead" – but to no avail.
If anyone is able to figure out how to solve this, I would be very grateful.
Thanks!
Best Answer
If an airplane is filled with $150$ passengers, at least one passenger gets bumped in the cases $k\in\{151,\ldots,160\}$. To get the total chance of at least one passenger missing the flight, one has to take a summation over these different chances to get something like $$\sum_{i=151}^{160}P(X=i;160,0.9)=\sum_{i=151}^{160}\binom{160}{i}\cdot0.9^i\cdot0.1^{(160-i)}.$$
Plugging this into wolfram alpha indeed gives a chance of $0.0359$, which corresponds to the one mentioned in the question. I seriously wouldn't recommend calculating this by hand, though.