If I'm doing it right, that does help, because there should only be a 19/27 chance of getting one red, a 6/27 chance of two red and a 1/27 chance of three red. I think I'm doing something wrong though, because that would only leave a 1/27 chance of NO red, and that should be twice as likely as ALL red since there's twice as many ways it can happen. This is where using card play doesn't help me; I can tell you the odds of a hand getting the queen of spades on a given split, and tell you the odds of that particular split, but I've never had to consider the odds of a person getting TWO queens of spades. ("DIRECTOR!!!" )

Typically I brute force these as a check when they're simple enough:

The possibilities for exactly one red are RBB, BRB, BBR (and the same for red and green, ) RBG, RGB, BRG, BGR, GRB and GBR, 12 total.

The possibilities for exactly two red are RRB, RBR, BRR, RRG, RGR and GRR, 6 total.

The possibility for exactly three red is RRR only, so 19 in all.

These same possibilities exist for cases of blue and green, but we've already counted all those containing red, leaving cases of BGG, GBG and GGB for exactly one blue (3), BBG, BGB and GBB for exactly two blue (3) and BBB as the only all blue (1). That leaves VERY few cases of green that we haven't counted; in fact, it just leaves GGG. So 19+7+1=27 possibilities, each of equal likelihood. That means there is a:

A) 12/27 chance of exactly one red one,

B) 6/27 (or 2/9) chance of exactly two red ones,

C) 1/27 chance of all red and

D) 8/27 chance of NO red.

It appears that when I treated cases of exactly one red as 2/3*2/3*1/3+2/3*1/3*1/3+1/3*1/3*1/3 I was actually counting the cases of ANY red, which is obvious in retrospect, but then I was still thinking in terms of "does 'one red' mean 'ANY red' or 'EXACTLY one red'?" If I'd thought about it a little more I'd have probably realized there are as many cases of exactly two blue/green as red, just as I realized there are as many cases of all blue/green (though that's a cumbersome way to tackle the first question. ) In itself, the question "what are the chances of two red?" has the same problem; "all red" might count, too, in which case there is 1 more possibility. The way YOU framed the question makes it clear; the way the lecturer framed it does not. I should've considered your questions together, because by taking them one at a time I did the first one one way and the rest a completely different way.

Sidebar: This is how I got through math tests in school; I invariably finished last, but I also invariably finished among the highest scores, solving problems, and solving them the right way, through sheer tenacity because I didn't let go until I had the right answer and KNEW it was right. Of course, if I misunderstood what the question initially asked I was still screwed because I could selfcheck for the wrong data all day long and it would always look right.

That's a lot longer than needed to simply answer the questions, but illustrates my thinking, just as you illustrated how to think about it properly rather than just stating the answer. Well done on that, by the way; you must be a pretty good teacher, and we could use more like you.

This gets a lot easier now that the process has been dissected; again, there are 2^3 possibilities, of which we've eliminated GGG, leaving 7 ways to get at least one boy. Knowing that, we're only dealing with two kids (but still 7 outcomes, not 4; that's important) so there are 2^2 possibilities, of which 1 is both girls, 2 are one of each and 1 is both boys, meaning a 2/7 chance of exactly one more boy and a 1/7 chance of all boys if you have one already. This can be checked as with the last example; possible outcomes are:

BBB, BBG, BGB, BGG, GBB, GBG, GGB and GGG.

All but 1 of those contains at least one boy, and of those 7, 2 have exactly two boys and 1 has three.

It's a little tricky in that when we say "you have three kids and one is a boy" we SEEM, at first glance, to be eliminating an entire child from consideration, but, really, we're only eliminating 1 outcome while still considering all three kids in the overall math.

I still think the biggest problem is that the question was vaguely worded; you can know how to do it right for both meanings and still get the desired answer wrong if you don't know what the questioner is actually asking.