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**kalyanram** This is incorrect the probability of Mrs. Brown is $\displaystyle \frac{3}{14}$ hence over all probability = $\displaystyle \frac{3}{14}.\frac{10}{13}.\frac{5}{12}$

The probability is with respect to the seat selection after two are occupied.

This argument is incorrect. The correct argument goes like this you have picked $\displaystyle 3$ suitable places for Mrs. Brown and $\displaystyle 10$ suitable places for Mrs.Lin. Now your job is to place one student in front of Mrs.Lin; and the choices you have are $\displaystyle \binom{5}{1}=5$ and the remaining people can pick the seats in $\displaystyle {^{11}P_9}$ ways. And as you have mentioned total possible arrangements are $\displaystyle {^{14}P_{12}}$. Probability of the event is $\displaystyle \frac{3.10.5.{^{11}P_9}}{{^{14}P_{12}}}$

Kalyan.