Suppose that, on the day an assignment is due, students hand in the assignment at an average rate of 4 per hour. Suppose the assignment is due at midnight, so there are 24 hours in which students can hand in the assignment without it being late.
(a) What is the probability that in the first student to hand it in takes at least 10 hours?
$$P(X\geq 10)=1-P(X\leq 9)=1-\sum_{k=0}^{9} \frac{e^{-1/4}(-1/4)^k}{k!}\approx 0.393$$
(b) What is the probability that it takes more than 3 hours for 20 students to hand in the assignment?
$$P(X>20)=1-P(X\leq 20)= 1-\frac{12^{20}e^{-12}}{20!}$$
(c) Suppose there are 100 students in the class. Calculate the probability that any student is one day late by modeling hand ins as a Poisson process using two expressions (one for P(event), one for 1-P(event’)). Why are these different (i.e., why is this an invalid approach to solving this problem)?
Not exactly sure about this one. What I have is:
$P(event)=\sum_{k=0}^{99} \frac{e^{-96}96^k}{k!}\approx 0.645$
$1-P(event)=(1-P(X\geq 24))^{100}=(1-e^{-6})^{100}\approx 0.78$
(d) Name one way in which this model is fitting for students handing in the assignment.
Not sure about this one.
(e) Name one way in which this model is ill-fitting for students handing in the assignment.
There's a fixed number of students but the Poisson distribution handles from 0 to $\infty$ number of students.
(f) What would a better model be, and how does that model overcome the limitations of this model?
Binomial is my guess since there are two mutually exclusive outcomes i.e. late or on time.
Are the answers I have given thus far correct? I'm still stuck on parts c, d, e and f
We need to be clear about what your random variables represent, and what is the model you are using. Based on your answers so far, it looks like you want $X$ to represent the random number of students that turn in the assignment in some fixed period of time. But you are totally unclear about the details. Moreover, you fail to distinguish between a time-to-event random variable, versus a counting random variable.
Let's set up the model correctly. We will let $X(t)$ represent the random number of students turning in the assignment over a time span of $t$ hours. So $X(4)$ represents the random number of students who turn in the assignment in the first four hours on the day it is due, and $X(24)$ represents the random number of students who turn in the assignment on time--i.e., at some point during the day it is due.
Then $$X(t) \sim \operatorname{Poisson}(\lambda t), \quad \lambda = 4, \\ \Pr[X(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!}, \quad x \in \{0, 1, 2, \ldots \};$$ that is to say, the number of such students is Poisson distributed with mean intensity $4t$. Why is $\lambda = 4$? This is because we are told that the average rate of submissions is $4$ per hour and $t$ is measured in hours.
But it is important to note that $X(t)$ is a counting process, i.e., it gives a number of students, hence must be a nonnegative integer; and not the time at which something happens, which could be any nonnegative real number. To understand how we model the latter, we might let $T_x$ represent the time it takes for the $x^{\rm th}$ student to turn in the assignment, so for example, $T_1$ is the time it takes for the first student to turn in the assignment; $T_2$ is the time it takes for the second student to turn in the assignment, etc. For convenience we define $T_0 = 0$, so that $T_x \ge T_{x-1}$ for every positive integer $x$, and $T_x - T_{x-1}$ is the $x^{\rm th}$ interarrival time. This is the random time elapsed between successive students turning in the assignment.
One important property of the (homogeneous) Poisson process is that $$T_x - T_{x-1} \sim \operatorname{Exponential}(\lambda), \\ \Pr[T_x - T_{x-1} \le t] = 1 - e^{-\lambda t}, \quad t > 0.$$ Interarrival times are independent and identically distributed exponential random variables with rate $\lambda$. So in particular, the random time for the first student to turn in the assignment is an exponential random variable with mean $1/4$ hour; hence the probability that it takes at least $10$ hours for the first student to turn in the assignment is $$\Pr[T_1 \ge 10] = e^{-10\lambda} = e^{-40} \approx 4.24835 \times 10^{-18}.$$ This is a tiny number and is consistent with our intuition: if on average $4$ students turn in the assignment every hour, the chance that nobody will turn in the assignment in the first $10$ hours is going to be extremely small.
Another way to perform the computation is to use the counting random variable. Note that the statement that it takes at least $10$ hours for the first student to turn in the assignment is equivalent to saying that in the first $10$ hours, no students turn in the assignment; i.e., $$\Pr[T_1 \ge 10] = \Pr[X(10) = 0].$$ Then we apply the Poisson PMF to obtain $$\Pr[X(10) = 0] = e^{-4(10)} \frac{(4(10))^0}{0!} = e^{-40}.$$
In general, it is not difficult to see that $$\Pr[T_x > t] = \Pr[X(t) < x]. \tag{1}$$ The left-hand side is the probability that the $x^{\rm th}$ submission takes more than $t$ hours to occur. The right-hand side is the probability that fewer than $x$ students turn in the assignment within the first $t$ hours. The two things mean the same thing, just expressed in different ways.
For the second part of your question, the event that it takes more than $3$ hours for $20$ students to turn in the assignment is $T_{20} > 3$, which we have shown above is equivalent to $X(3) < 20$ (fewer than $20$ students turn in the assignment in the first $3$ hours). Thus you want to compute $$\Pr[X(3) < 20] = \Pr[X(3) \le 19] = \sum_{x=0}^{19} e^{-4(3)} \frac{(4(3))^x}{x!}.$$
For part (c), the question is poorly phrased, so the only thing I can do is explain how to perform the correct calculation. The probability that at least one student is late in a total class size of $n = 100$ students is equal to the probability that there are fewer than $100$ submissions in $24$ hours; i.e., it is $$\Pr[X(24) < 100] = \Pr[X(24) \le 99] = \sum_{x=0}^{99} e^{-4(24)} \frac{(4(24))^x}{x!}.$$ I don't understand how an incorrect calculation would be set up.
For (d), I would say that if there are a sufficiently large number of students in the class, the Poisson model would be adequate, since it satisfies certain convenient properties. For instance, if students work independently of each other, their submission times are independent.
For (e), you are correct. However, I would also add that the Poisson model would not fit well even if there are infinitely many students, because in the real world, students do have a tendency to wait until the last moment to turn in their work; they also tend not to submit homework during certain hours of the day.
The last question is vague because "better" depends on the ways in which we quantify "better." For instance, as you pointed out, if there are a finite number of students, maybe even a very small number--say $10$ students--a better model for the number of students who submit the assignment on time would be binomial, assuming that all we wanted to know is how many students are on time, and we are able to assign a Bernoulli probability to each student's chance of being on time. But the binomial model does not model when each student turns in the assignment (this is an advantage of the Poisson process).