How to find the expected number of days? Here is the problem that I am trying to solve:
A hospital handles 20 births a day. Ten percent require a special fetal monitor. Find the expected number of days out of the year when the hospital will need at least two monitors.
Any help would be great.
 A: As also pointed out in the comments, we are assuming that the number of samples is so large such that the probability that each newborn needs the device is $0.1$, independent of others.
Assume any day. The probability that no newborn needs the monitor is $0.9^{20}$. Similarly, the probability that exactly one monitor is needed is $(20)(0.1)0.9^{19}$. Hence the probability that at least two monitors are needed becomes $p=1-0.9^{20}-(20)(0.1)0.9^{19}=0.6083$.
To calculate the expectation, notice that the number of days that the desired event happen follows a binomial $n=365,p$ distribution. 
Therefore, the expected number of days that this event happens is $365p=222.3$ days.
A: The answer made by msm is good. Just with a small mistake: the probability that exactly one monitor needed should be $(20)(0.1)0.9^{19}$. Thus, the probability that at least two monitors are needed becomes $p=1−(0.9)^{20}−(20)(0.1)(0.9)^{19}=0.6083$.
Taking $365.25$ days to be the expected number of days a year, the expected number of days out of a year that the hospital needs at least two monitors is $365.25 p = 222$ days.
A: You need the probability of needing at least two monitors on a given day. Once you have that, multiply by the number of days in a year to get the answer.
To get the probability of needing at least two or more monitors, think about it like this: Each birth is like a weighted coin flip that comes up "heads" (i.e. patient needs a monitor) only 10% of the time. So you need the probability of having two or more "heads" out of 20 biased coin flips under these conditions. The binomial distribution will come in handy.
A: if the 20 births a day is an average, then you could use poisson with $ \lambda = 20 \times 0.1 = 2$
$P(0) = e^{-\lambda} =$
$P(1) = \lambda e^{-\lambda}$
$P(>=2) = 1 - P(0) - P(1) = 1 - .1353 - .2707 = .594
no of days per year = .594 x 365.25  = 217
