Probability of arranging students into groups

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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.

Re: Probability of arranging students into groups

If A_1 **and** A_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).

Quote:

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

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.

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.

Re: Probability of arranging students into groups