# could youplease tell me what is wrong in my solution?

by abhishek   Last Updated July 12, 2019 08:20 AM

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 $$P(N = 0) \approx \frac{e^{-1}1^0}{0!} = e^{-1}$$

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

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