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:

$\displaystyle P(X = 150; 160, 0.9) = \binom {160}{10} * 0.9^{150} * 0.1^{10}$


$\displaystyle \frac{160*159*158*157*156*155*154*153*152*151}{1*2*3*4*5*6*7*8*9*10} * 0.9^{150} * 0.1^{10}$


$= 0.03112979$

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.



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.

  • $\begingroup$ Yeah, now that would take probably two pages to calculate by hand... Okay though, I think I understand it, could you confirm? Essentially, by saying (in the context of a binomial distribution) that P(X > k) where n is 160, they essentially want the sum of all numbers where it is greater than k(150), but less than or equals n(160)? $\endgroup$ – Brugsen May 20 '14 at 7:46
  • $\begingroup$ and this could be deduced by the question "What is the probability that at least one passenger will have to be ʻbumpedʼ from the flight?" implies > 150 passengers show up... So is what he sum is doing essentially calculating the chance that 1 extra shows up, 2 extra shows up, 3 extra shows up... up to 10 extra show up? If the question was at MOST one passenger to be bumped, would it essentially be $\displaystyle P(X = 151; 160, 0.9)$ - or would it be the sum of 1 to 151? Thanks. $\endgroup$ – Brugsen May 20 '14 at 7:51
  • $\begingroup$ Can't edit, but at most 1 would be 0 or 1, so 150, and 151? $\endgroup$ – Brugsen May 20 '14 at 8:54
  • $\begingroup$ That sounds about right, indeed. "At most one passenger" would be interesting. It sounds like you'd really need the sum of 0 to 151: if the plane is (nearly) empty, no passenger would get bumped, would they? In that case it's easier/quicker to calculate 1-(sum of chances for $k>151$). The chance of having to bump exactly one passenger would correspond to $k=151$. $\endgroup$ – HSN May 20 '14 at 11:39

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