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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 $P(N = 100) \approx \frac{e^{-99} 99^{100}}{100!} = 0.03966085721 \neq e^{-1}$

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    $\begingroup$ The Poisson approximation to the Binomial distribution is only valid when $p$ is very small because from $\lambda=np$ we have $p=\frac{\lambda}n$ and we need $n$ to be large. See also here $\endgroup$ – Peter Foreman Jul 12 at 8:30
  • $\begingroup$ that was helpful, thanks! $\endgroup$ – abhishek Jul 12 at 8:41

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