Poisson approximation to the Binomial distribution - verify solution

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

• 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 – Peter Foreman Jul 12 at 8:30
• that was helpful, thanks! – abhishek Jul 12 at 8:41