Triplets Adam, Bruce, and Charlie enter a triathlon.

[ch339]

Triplets Adam, Bruce, and Charlie enter a triathlon. There are nine competitors in the triathlon. If every competitor has an equal chance of winning, and three medals will be awarded, what is the probability that at least two of the triplets will win a medal?

A) 3/14
B) 19/84
C) 11/42
D) 15/28
E) 3/4

Source: Advanced Quant Workout Set 1, Question 4

So I get that there are 9C3 ways to assign winners. There are 3C3 +3C2 ways that at least two triplets could win. I got 4/84. I see now that we have to take the full event into account (i.e., 2 brothers and someone else, so 3C2*6C1= 3*6 --> (18+1)/84).

But if we toss a fair coin 5 times and want exactly 3 heads, we get 5C3 = 10. So 10/2^5 = 10/32. Or at least 3 heads, (5C3+5C4+5C5)/32 = (10+5+1)/32 = 1/2. Why don't we say 3 heads and 2 tails (5C3*5C2 = 100 --> 100/32)? I understand that it is impossible to have probability greater than 1, but how are the two problems (bros and coins) qualitatively different?

[RonPurewal]

If I were looking at this problem, I wouldn't bother with calculating anything. Because, well, I'm lazy and don't like to calculate things if I don't have to.

It's easy enough to just list all of the ways in which the criterion can be satisfied:
A, B, and C (1 way)
A, B, and _, where "_" is anyone else who is not C (6 ways)
A, C, and _, where "_" is anyone else who is not B (6 ways)
B, C, and _, where "_" is anyone else who is not A (6 ways)

That's 19 ways to do it.

So, the probability is 19 out of ... something.

There's only one answer choice that is 19/whatever, so, that choice wins. Why do unnecessary work?

It amazes me how many people are willing to do TONS of work on combinatorics problems"”but are completely unwilling to just make a list of possibilities.
Sometimes the best solution is just "pick up the shovel, and shovel the dirt".

[RonPurewal]

In terms of this:

But if we toss a fair coin 5 times and want exactly 3 heads, we get 5C3 = 10. So 10/2^5 = 10/32. Or at least 3 heads, (5C3+5C4+5C5)/32 = (10+5+1)/32 = 1/2. Why don't we say 3 heads and 2 tails (5C3*5C2 = 100 --> 100/32)? I understand that it is impossible to have probability greater than 1, but how are the two problems (bros and coins) qualitatively different?

They're not different. They're exactly the same.

The difference is that you're making two completely different selections: two of the three brothers, and one of the six non-brothers.
So, if you use the same (incorrect) way that gives you 5c3 * 5c2, then you'll get (3c2 * 3c1)(6c1 * 6c5), which is likewise incorrect.

[RonPurewal]

By the way, it makes me cringe to see "6c1""”or, for that matter, anything"c1""”in a work-up, since it's just so ... unnecessary. Moreover, it reveals that the person doing the work isn't thinking intuitively.

E.g., if you have to pick one person out of six, then there are obviously six ways to do so. If someone needs a formula to do that ... not good.

[ch339]

But how is 3 H and 2 T different from 2 triplets and 1 non-sibling?
If you are choosing exactly 2 triplets, why does it matter? These problems confuse me.

I meant 3C2*6C1 + 3C3. But, I think I get it now. We have to choose groups of three, and 2 bros & 1 non-bro is a triad.

And yes, it's just easier to type 6C1 = 6 than there are six ways to ... I realize there are 6 ways to pick one person and one way to pick all three for a group of three. Thanks for all your help!

[RonPurewal]

You can frame it as a choice of 3 heads and then 2 tails, if you want to.
But, if you do that, the second expression is not 5c2. It's 2c2 (which is just 1, and so goes away).

I.e., after you've chosen which three tosses are heads, you can't pick two tails out of five tosses anymore.
Once the three heads have been chosen, there are only two remaining tosses, both of which must be tails. You can regard this as "choosing 2 tails out of the 2 remaining tosses" (2c2) if you want, although I think it's evident why that is unnecessary.

There's no parallel to the other problem. In that problem, the non-sibling medal winner is still chosen from the six that remain.
If you want to cast that problem in the same way you're casting this one, you'd have to do this:
* Choose 2 siblings (of 3) to win medals;
* Choose 1 non-sibling (of 6) to win a medal;
* Choose 1 sibling (of the remaining 1) to NOT win a medal;
* Choose 5 non-siblings (of the remaining 5) to NOT win medals.
That's (3c2)(6c1)(1c1)(5c5), which is just (3c2)(6)(1)(1). same as the existing solution.

Hope that makes things more clear.

[RonPurewal]

So, just to be clear, your approach (considering the heads and then the tails) gives (5c3)(2c2) = 5c3 for the first problem. This agrees with the existing solution.

Make sure the importance of my first post isn't lost on you. If you have ANY doubts AT ALL about how to evaluate a combinatorial quantity ... just MAKE A LIST.

As I wrote above, I wouldn't mess with combinatorial calculations at all here. I'd just write out the possibilities. That way it's absolutely guaranteed that the answer is right, since I'm staring at an exhaustive list and can just count things.
I like just counting things. Simple makes me happy.

[ch339]

Yes, makes sense. I was confused initially because I thought we only cared about the the 3 spots (winners and heads) and I could just ignore the non-siblings as I would tails. I know 3 heads = 5C3, but I ignore the tails since choosing exactly 3 heads is the same as choosing exactly 2 tails. I mistakenly thought that if I wanted to choose exactly 2 brothers, this would just be 3C2 = 3C1 = 3. But that ignores the third person in the group, which is not the same as ignoring the irrelevant other 6 positions (analogous to the 2 tails). You need exactly 2 bros AND a non-sibling to complete the triad. So there are 3*6 ways to do this. I will try listing in the future as well. I find my understanding of combinatorics problems is either hit or miss. Thank you again for your detailed explanations.

Simple makes me happy, too!

[RonPurewal]

You're welcome.