Probability Problem (Poisson Distribution) It has been observed that statistics majors in a certain university produce $60$ hypothesis test results, on average, daily. What is the probability that a particular statistics major from that university
$a$) will produce at least one hypothesis test results tomorrow
$b$) produces $70$ hypothesis test results in two days
My solution at $a$ is ($e^{-60}$$(60)^0$ ) / $0!$ and at $b$ is ($e^{-120}$$(120)^{70}$) / $70!$ . Are my answers correct or I misinterpret the said problem?
 A: Putting details of this somewhat fanciful 'homework overload' situation aside, let's look
at your attempt to apply the Poisson distribution:
In (a), suppose $X \sim \mathsf{Pois}(60),$ and that you seek
$P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-60}.$ Your result is correct.
The numerical answer is very nearly 1, because $P(X=0) \approx 0.$
In (b), suppose $Y \sim \mathsf{Pois}(120),$ and that you seek
$P(Y = 70) = e^{-120}(120)^{70}/70!.$ Your result is correct.
Specifically, you correctly doubled the Poisson rate to match the 2-day time period. (Note: If you try to compute $70!$ on a typical calculator,
it will overflow the capacity of the internal software because
$70! > 10^{100}.$)
If the problem intended $P(Y \ge 70),$ then the answer is much
larger. You could use software for an exact answer or a normal 
approximation (with $\mu = 70$ and $\sigma = \sqrt{70}).$
Using R statistical software, one gets the following numerical results.
1 - dpois(0, 60)                 # (a)
## 1
dpois(70, 120)                   # (b) P(Y=70)
## 2.233286e-07
1 - ppois(69, 120)               # (b) P(Y >= 70) exact
## 0.9999997
1 - pnorm(69.5, 120, sqrt(120))  # (b) P(Y >= 70) norm aprx w/ continuity correction
## 0.999998
factorial(70)                    # 70! is HUGE
## 1.197857e+100

