Poisson approximation to the Binomial distribution - verify solution

434 Views Asked by At

Harvard Law School courses often have assigned seating to facilitate the “Socratic method.” Suppose that there are 100 first year Harvard Law students, and each takes two courses: Torts and Contracts.
Both are held in the same lecture hall (which has 100 seats), and the seating is uniformly random and independent for the two courses.
Find a simple but accurate approximation to the probability that no one has the same seat for both courses.

given solution:

Define $I_i$ to be the indicator for student $i$ having the same seat in both courses,so that $N=\sum_{i=1}^{100}I_i$ , Then $P(I_i = 1) = 1/100$, and the $I_i$ are weakly dependent.
So $N$ is close to $Pois(\lambda)$ in distribution, where $\lambda = E(N) = 100E[I_1] = 1$. Thus $P(N = 0) \approx \frac{e^{-1}1^0}{0!} = e^{-1}=0.36787944117 $

my solution:

Define $I_i$ to be the indicator for student $i$ not having the same seat in both courses,so that $N=\sum_{i=1}^{100}I_i$ , Then $P(I_i = 1) = 99/100$, and the $I_i$ are weakly dependent.
So $N$ is close to $Pois(\lambda)$ in distribution, where $\lambda = E(N) = 100E[I_1] = 99$. Thus $P(N = 100) \approx \frac{e^{-99} 99^{100}}{100!} = 0.03966085721 \neq e^{-1}$