Question about Poisson processes I'm given that during some 30-day month (say, November), a chime randomly rings according to a Poisson process with a rate of one chime every six days.
Letting $A$ represent the number of days in this month that will have at least one chime, I'm trying to find $E(A)$.
Would $E(A)$ just equal $\lambda = {30\over6} = 5$?
Or would I have to solve for:
$$P(N(1)\ge 1) = 1 - P(N(1)=0)$$
$$= 1 - {({1\over6})^0*e^{-{1\over6}}\over{0!}}$$
$$E(A) = n*p=30*p\approx4.606$$
... where I have $\lambda = {1\over6}$ and $t=1$.
 A: The correct approach is the second, due to the following reasoning.
The number of days with at least one chime (success) $A$ in a time period of 30 days is a binomial random variable with paremeters $n=30$ and $p$ given by $$p=P(N(1)\ge1)$$ where $N(1)\sim \mathrm{Poisson} (λ=1/6)$, exactly as you have it. So, the expected value of $A$ is $$E[A]=np$$ as your second approach yields.
Your first approach gives "the expected number of chimes in this month" (which is 5, as you correctly calculated) and not "the expected number of days with more than one chime in this month".
A: The expected number of chimes would be $1/6$ in a day, thus the formula for the probability of $k$ chimes in a given day would be
$$
\frac1{6^kk!}e^{-1/6}
$$
The probability of $0$ chimes in a given day would be $e^{-1/6}$, so the probability of getting at least one chime on a given day would be $1-e^{-1/6}$. By the linearity of expectation, we should expect to see
$$
\mathrm{E}(A)=30\left(1-e^{-1/6}\right)\doteq 4.605548
$$
days with at least one chime.
A: $30/6$ is the total expected number of chimes over 30 days, which is not what you want because some days can have multiple chimes. So your answer should be less than $5$. The second approach you outlined is the right way to go, and gives the right answer by linearity of expectation.
