# binomial distribution(overbooking plane tickets)

I am having trouble with binomial distribution and this problem: an airplane has 200 seats, but 202 tickets are sold. Assume passengers do not show up with a probability of .03 independently. What is the chance that a flight is over full? Let X be the random variable that equals 1, then P(X=1)=P(PPPP...PM in any order) where P means passenger is present and M means passenger is missing. To do this we find how many ways are there to get this sequence of successes/failure(missing/present passengers):(202 choose 1)(.97)^201*(.03)^1=.01329 chance of a flight being overbooked? is that right or do i have to sum P{X=1,0}=P(X=1)+P(X=0)? I'm also confused at (202 choose 1)(.97)^201*(.03)^1, shouldn't that equal .03? We are saying out of the 202 tickets choose 1(in 202 different ways) to have one person missing at a probability of .03 and all other present at a probability of .97. Shouldn't that probability be .03?

• You do have to deal separately with what you call $1$, and $0$ and add up. – André Nicolas Nov 18 '13 at 7:11
• editing my post, i made some errors. It should have been 1 and 0 instead of 2,1,0 since there are 202 tickets sold and overbooking would mean 201 tickets. Can you explain why is my logic wrong in thinking that (202 choose 1)(.97)^201*(.03)^1 should equal .03? If instead of tickets we had a flip of a coin and we wanted to find the probability of getting a head with a tail being a probability of .50 independently then wouldn't it be (1-.50)^0*(.50) which is .50 or a 1/2. – bob the builder Nov 18 '13 at 7:12
• This seems like the same concept, flipping a coin and getting a head with a .50 probability of a tail vs getting a missing passenger with a .03 probability rate. P{Head} at 1/2 or P{Tail} at 1/2 vs P{Missing} at .03 or P{Passenger} at .97 – bob the builder Nov 18 '13 at 7:20
• It is the same concept, and the expression is similar, apart from the simplification from the fact that $1-0.5=0.5$. If we flip a fair coin $12$ times, the probability of exactly $11$ heads is $\binom{12}{11}(0.5)^{11}(0.5)^1$. – André Nicolas Nov 18 '13 at 7:25

Let random variable $X$ be the number of people who show up. The probability that the plane is overbooked is $$\Pr(X=201)+\Pr(X=202).$$

We have $$\Pr(X=k)=\binom{202}{k}(0.97)^k (0.03)^{202-k}.$$

Alternately, we can calculate the two needed probabilities without explicit appeal to the formula for $\Pr(X=k)$.

The probability that $202$ people show up is $(0.97)^{202}$.

For the probability that $201$ people show up, the missing person can be chosen in $202$ ways. The probability she doesn't show up is $0.03$. The probability all the others show up is $(0.97)^{201}$, so the probability $201$ people (exactly) show up is $(202)(0.03)(0.97)^{201}$.

Remark: Equivalently, we can let $Y$ be the number of missing passengers, and find $\Pr(Y=0)+\Pr(Y=1)$. That seems to be what you chose, though the random variable was not defined precisely.