Conditional Probability

savary wrote:A class consisting for 4 graduate and 12 undergraduate students is randomly divided into 4 groups of 4. What is the probability that each group includes a graduate student.

Let the 4 grad students be A, B, C and D.

Among the 4 groups of 4 are a total of 16 positions.
Student A can occupy any of the 16 positions.
.
Of the 15 remaining positions, 12 are not in A's group.
Thus:
P(B is not in A's group) = 12/15.

Of the 14 remaining positions, 8 are not in A's group or B's group.
Thus:
P(C is not in A's group or B's group) = 8/14.

Of the 13 remaining positions, 4 are not in A's group, B's group, or C's group.
Thus:
P(D is not in A's group, B's group, or C's group) = 4/13.

Since we want all of these events to happen, we MULTIPLY the probabilities:
12/15 * 8/14 * 4/13 = 64/455.

Thank you Mitch, this was very helpful.

I have this book on my shelf! (I can see why the worked example didn't help; the problem sets are great, but Bertsekas himself can be terse and tough to follow.)

Let me add a combinatorial approach, since this generalizes well to other problems you'll encounter later in real probability.

When we go to pick the first group, we want ONE grad and THREE undergrads. This can happen in 4*(12 choose 3) ways, since we're choosing one of four grads and three of twelve undergrads. This gives Target / Total = 4*(12 choose 3) / (16 choose 4). (The total is just all the ways we could randomly pick four people of the sixteen.)

When we go to pick the second group, we have 12 students left, and again want ONE grad and THREE undergrads. This can happen in 3*(9 choose 3) ways. This gives Target / Total = 3*(9 choose 3) / (12 choose 4)

When we go to pick the third group, similar logic gives us 2*(6 choose 3) ways. This gives Target / Total = 2*(6 choose 3) / (8 choose 4).

Since the fourth group is now predetermined, we don't have to bother with it.

Multiplying all the probabilities above gives us

4 * 3 * 2 * (12! / 3!9!) * (9! / 3!6!) * (6! / 3!3!) / ((16! / 4!12!) * (12! / 4!8!) * (8! / 4!4!))

As you can see when you write it out, this cleans up much more nicely than you'd expect, and leaves 64/455.

Matt, I'm sorry I didn't read your post earlier. I was going through the problems again after studying combinatorics. Actually, I was using the book along with video lectures from MIT OCW course entitled "Probability Models and Axioms". Having done the problem both ways, I found the combinatorial approach easier to follow but more time-consuming in calculations.