I am not quite sure of how the book did this question. I did it another way using counting.

Consider students 1, 2, 3, 4 are all in different groups, then there are 12!/(3!)^4 different ways that the other 12 students can be put into groups.

There are 16!/((4!)^4*4!) to put 16 students into 4 groups of 4, the *4! accounting for the overcounting of the order of the groups.

The final probability follows.

However with the book's working, how is P(A_3) = P(A_1 n A_2 n A_3)?

Isn't A_1 n A_2 = A_1?

Since (student 1,2 in different groups) n (student 1, 2, 3 in different groups) = (student 1,2 in different groups) ?

Also how did they get P(A_1) = 12/15?

Like I just don't understand the intuition of how they arrived at the probability.

Thanks.