Using conditional probability in a tricky way I was going through the problem attached herewith. the fact is I could not understand 2 things. Firstly, how did the author came up with the idea of choosing the event sets the way he has described? How is this intuitive? Secondly, I tried to find the $P(A_1)$ directly, and I came up with the following way. $P(A_1)=(C(4,2)\cdot 2!\cdot(14!/(4!^2\cdot3!^2)))/(16!/(4!^4))=12/15$. I cannot intuitively follow the procedure given here. Any help? ![enter image description here](

It may become more intuitive if you introduce event $A0$ as student $1$ occupying any box.

Students $2,3,4$ have to successively occupy a slot from the still **unoccupied group(s)** of $4$ slots

$A0:\;\square\square\fbox{1} \square \quad\square\square \square\square\quad\square\square\square \square\quad\square\square\square \square\quad Pr = 1 $

$A1:\;\square\square\fbox{1} \square \quad\square\square \square\square\quad\fbox{2}\square\square\square\quad\square\square\square \square\quad Pr = 1\cdot \frac{12}{15} $

$A2:\;\square\square\fbox{1} \square \quad\square\square \square\square\quad\fbox{2}\square\square \square\quad\square\square\square \fbox{3}\quad Pr = 1\cdot \frac{12}{15}\cdot\frac8{14} $

$A3: \;\square\square\fbox{1}\square\quad\square\fbox{4}\square\square\quad\fbox{2}\square\square\square\quad\square\square\square \fbox{3}\quad Pr = 1\cdot \frac{12}{15}\cdot\frac8{14}\cdot\frac4{13} $

We are simply using the multiplication law for dependent events.