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The probability that at least two of the triplets will win a medal is the sum of the probability that exactly two of the triplets will win a medal and the probability that all three will win a medal.

The probability that exactly two of the triplets will win a medal is \frac{C^2_3*C^1_6}{C^3_9}=\frac{18}{84}, where C^2_3 is ways to select which two of the triplets will win a medal, C^1_6 is ways to select third medal winner out of the remaining 6 competitors and C^3_9 is total ways to select 3 winners out of 9;

The probability that all three will win a medal is \frac{C^3_3}{C^3_9}=\frac{1}{84};

P=\frac{18}{84}+\frac{1}{84}=\frac{19}{84}.

Answer: B.

Hope it's clear.

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From what I understand in the question is that there are 3 teams (triplets) of 3 people each competing together , so how is it possible that only 2 triplets/ teams win medals as there would be a definite 1,2,3 position in case of 3 teams. ?? ---. because the word triathlon implies to me that the teams win not the individuals (teams with cumulative individual lowest score times win) ....

Now if we consider this otherwise that individuals win then :

The probability that at least two of the triplets will win a medal is the sum of the probability that exactly two of the individuals from same team/triplet wins the medal and third medal is won by a individual from one of the remaining 2 teams/triplets + 1 Individual from each team/triplet wins the medal...

P(at least two of the triplets will win a medal is the sum of the probability that exactly two of the individuals from same team/triplet wins the medal and third medal is won by a individual from one of the remaining 2 teams/triplets)=3P2 * 3/9 * 2/8 * 6/7 = 3/7

P(1 Individual from each team/triplet wins the medal)= 3P3 * 3/9 * 3/8 * 3/7 = 9/28

3/7 + 9/28 = 3/4 ---- (E)

Pl. help me know where I am going wrong.
Thanks.

I might be combinatorics challenged...
I got that for 2 of the triplets to get award and one other competitor -
probability = 3C2 * 6C1 / 9C3 = 3!*6!*3!*6! / 2!*5!*9! = 3*6*3 / 9*8*7 = 3/28

mCn = m!/(n!*(m-n)!)

How is it that you got 18/84 ???

Hi Bunuel

Isnt this a permutation problem? I did in the same way as you did except that I used permutation since I thought that the rank and the order in which they are acquired does matter here. Obviously I am wrong since its not the official answer.. What am I missing here??

It is a combinatorics question and not a permutation because it asks of two getting a medal (placing either 1st 2nd or 3rd) but does not consider the order.

As a result there are 3 ways in which the podium ( podium is a term used for the top three finishers in a race) can have 2 of the triplets: A and B ; A and C ; B and C.
There is only one way in which all three can occupy the podium: A, B, and C. Their finishing position does not count, simply that they made the podium.

It is not a Permutation, it is a combinatoric problem. So the formula is \frac{n!}{k!(n-k)!}

Even without the formula, if you think about what your list shows, you have doubled some selections. A and B is the same as B and A

Think of it this way: Treat the podium as a cup. If you put one A in the cup and one B in the cup, it is no different than one B and one A.

It's been quite some time since this was posted, but I have a quick question about how to solve this problem using fractional probability instead of combinatorics. I tried to solve the problem using the following:

P(\geq{\text{2 Triplets}})=P(\text{2 Triplets})+P(\text{3 Triplets})

P(\geq{ \text{2 Triplets}})=(\frac{3}{9} \times \frac{2}{8} \times \frac{6}{7})+(\frac{3}{9} \times \frac{2}{8} \times \frac{1}{7})

P(\geq{ \text{2 Triplets}})=\frac{6}{84}+\frac{1}{84}=\frac{7}{84}

I realize that in order to get to the original answer, I would have had to multiply the initial \frac{6}{84} by 3, and proceed to calculate \frac{19}{84}, but I do not understand why this is the case, since the order of the triplets should not matter.

In another problem from MGMAT's CAT test, I was able to solve for the desired probability using fractional probabilities only:

"Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?"

P(\geq \text{1 pair})=1- \text{ P}( \text{No Pairs})

P(\geq \text{1 pair})=1-(\frac{12}{12} \times \frac{10}{11} \times \frac{8}{10} \times \frac{6}{9})=1-\frac{16}{33}=\frac{17}{33}

Of course, this problem could have also been solved using combinatorics, but I feel the fractional approach would probably be quicker in a live scenario. However, over the course of my review, I've found the fractional method to be unreliable, whereas combinatorics are typically universally applicable (though they can take significantly longer in some instances).

I am curious as to why the solution for the second problem involving the playing cards worked whereas my solution to the original triplet question did not. I am not really sure why I have to multiply that initial \frac{6}{84} by 3 to get the right answer, and I would like to have a concrete understanding of this issue to help me solve similar problems.

DD, thanks for the quick response.

I'm still a little fuzzy on why \frac{3}{9} \times \frac{2}{8} \times \frac{6}{7} from the triplets problem implies that order matters whereas the \frac{12}{12} \times \frac{10}{11} \times \frac{8}{10} \times \frac{6}{9} from the card pairs problem does not.

Could you help explain this as well?