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?
[Math] binomial distribution(overbooking plane tickets)
binomial-coefficientsdiscrete mathematicsprobabilityprobability distributionsprobability theory
Best Answer
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 $(202)(0.03)(0.97)^{201}$.
Add up our two probabilities.
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.