If A_1andA_2, then (1&2 are in different groups)and(1&2&3 are in different groups), which is equivalent to the latter statement alone. Therefore, P(A_3) = P(A_1 n A_2 n A_3).

Put student 1 in a random group. Now there are 15 open slots, 3 of them are in the same group, 12 in others. Therefore, student 2 has a 12/15 probability to join a different group: P(A_1)=12/15.Also how did they get P(A_1) = 12/15?

In the same way, assume that student 1 and 2 are in different groups (A_1). Now there are 14 slots left, 8 in different groups. Student 3 has a 8/14 probability to join a new group: P(A_2|A_1)=8/14

Assume that students 1&2&3 are in different groups. The probability that student 4 will join the 4th group is 4/13: P(A_3|A_2)=P(A_3|A_2 and A_1)=4/13

Total probability: P=12/15*8/14*4/13=64/455.

I get the same number with your approach.