Your calculation of (a) is correct, but you should be more precise with your notation. Let $Y_t$ be the random number of calls that arrive in the next $t$ minutes. If $Y_t$ is a Poisson process with intensity/rate $\lambda$ per minute, then $$\Pr[Y_2 = 5] = e^{-\lambda t} \frac{(\lambda t)^5}{5!} = e^{-4} \frac{4^5}{5!}.$$
Your calculation of (b) is incorrect. You cannot add the intervals together to get $t = 4$, because when you do, you will no longer account for the fact that exactly $5$ events must occur within each $2$-minute interval. Instead, you must use the respective probability $\Pr[Y_2 = 5]$ and due to the independent increments property, you simply square this value to get the desired probability. In other words, if the probability that exactly $5$ events are observed in $2$ minutes is $p$, then the probability that you observe $5$ events between times $t = 0$ and $t = 2$ and another $5$ events between $t = 2$ and $t = 4$ is simply $p^2$. It would be wrong to compute $\Pr[Y_4 = 10]$, since this probability includes events where, for example, $3$ events occur in the first half and $7$ in the second half of the $4$-minute interval.
Your calculation of part (c) is correct.
As @DavidG.Stork Comments, you can ignore what happened
before 12:30, as long as you're not using that information to estimate $\lambda.$
Therefore, let $X \sim \mathsf{Pois}(\lambda = 90),$ where the
rate $\lambda = 90$ = (30 min)(3/min). Then
you want $P(X \ge 60) = 0.99967.$
You can get this exact answer using
R, where ppois is a Poisson CDF, as shown below. Some statistical calculators could do essentially the same
omputation.
1 - ppois(59, 90)
[1] 0.9996747
You might try a normal approximation to this Poisson
distribution, $\mathsf{Norm}(\mu = 90, \sigma=\sqrt{90}),$
standardize, and use printed tables of CDF of standard normal to get a reasonable normal approximation (with continuity correction).
The normal
approximation from R, where pnorm is a normal CDF, as shown below:
1 - pnorm(59.5, 90, sqrt(90))
[1] 0.9993477
Using normal tables you would get somewhat less accurate
version of this approximation, because some rounding error is involved in using such a table.
The figure below, compares $\mathsf{Pois}(\lambda=90),$ centers of red circles, with the density function of
$\mathsf{Norm}(\mu=90, \sigma=\sqrt{90}).$
R code for figure:
curve(dnorm(x, 90, sqrt(90)), 0, 140, lwd=2, ylab="PDF", main="")
abline(v=0, col="green2")
abline(h=0, col="green2")
k = 0:140; pdf=dpois(k, 90)
points(k, pdf, col="red")
abline(v = 59.5, col="blue", lwd=2, lty="dotted")
Best Answer
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X \sim Poisson(\lambda = 1.1)$ and $P(X=0) = e^{-1.1}$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $\frac{1}{\lambda} = 27.27$ so $\lambda = .0366$ and then $P(Y > 30) = 1-P(Y \le 30) = 1-(1-e^{-30 *.0366}) = e^{-1.1}$
I think the key thing is that $\lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.