Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 co

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.

Your understanding of the question is not correct. Adam, Bruce, and Charlie are not in one team, they compete between each other and 6 other competitors.

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.

Hey wfa,

You're right about order not mattering but your math suggests you're missing a fundamental counting concept.

P(2 triplets) is not simply equal to 3/9*2/8*6/7 as this would suggest a sequence where order matters. Rather, the math is (number of successful outcomes)/(total number of outcomes)

Total # of ways you could pick 2 of the triplets among the medallists is C(2,3), AND total # ways you could pick 1 non-triplet among the medallists is C(1,6), that's 3*6= 18

Total # outcomes is the # of ways you could pick ANYONE to be medallists, so C(6,9) = 84

P(2 triplets) = 18/84

I know u we're off by a factor of 3, and you think it's an order vs no-order issue, but really, you were attempting sequential counting, but should have been using combinatorials

Hope this clarifies things

Are Adam, Bruce and Charlie in the same triathlon team or are they in separate teams?

No, they are competing each other, they do not form a team.

Hi,

Note :- all medals are equal that is not different for 1,2 and 3 POSITION

Two cases
1) only two get medals..
Choose 2 out of three A,B, and C---3C2=3..
The third can be any of the remaining 6..
So total ways=3*6=18..
2) all three get medals..
Only 1 way..

Total ways ATLEAST 2 get medals= 18+1=19..

TOTAL ways=9C3=\(\frac{9!}{6!3!}=84\)..

Probability=\(\frac{19}{84}\)
B

\(\left\{ \matrix{\\
\,A\,,B\,,\,{\kern 1pt} C\,\,\,\,\, \to \,\,\,3\,\,{\rm{competitors}} \hfill \cr \\
\,{\rm{D,}}\,\, \ldots \,\,{\rm{,}}\,I\,\,\, \to \,\,\,{\rm{other}}\,\,6\,\,{\rm{competitors}}\, \hfill \cr} \right.\)

\(\left( {\rm{i}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}\)

\(\left( {{\rm{ii}}} \right)\,\,A,B,C\,\,{\rm{win}}\,\,{\rm{exactly}}\,\,2\,\,{\rm{of}}\,\,{\rm{the}}\,\,3\,\,{\rm{medals}}\)


\(? = P\left( {{\rm{i}}\,\,{\rm{or}}\,\,{\rm{ii}}} \right) = P\left( {\rm{i}} \right) + P\left( {{\rm{ii}}} \right)\,\,\,\,\,\,\,\,\left[ {\,{\rm{i}}\,\,{\rm{and}}\,\,{\rm{ii}}\,\,{\rm{are}}\,\,{\rm{mutually}}\,\,{\rm{exclusive}}\,} \right]\)


\(\left( {\rm{i}} \right)\,\,\,\left\{ \matrix{\\
\,{\rm{favorable}} = \,\,3! \hfill \cr \\
\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,P\left( {\rm{i}} \right) = {1 \over {84}}\)


\(\left( {{\rm{ii}}} \right)\,\,\,\left\{ \matrix{\\
\,{\rm{favorable}} = \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,A,B}\,\,\, \cdot \,\,\underbrace {C\left( {3,2} \right)}_{{\rm{say}}\,\,\,1{\rm{st}}\,\,,\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\underbrace {2!}_{{\rm{say}}\,\,\,A\,\,1{\rm{st}}\,\,,\,B\,\,2{\rm{nd}}}\,\,\, \cdot \,\,\,\underbrace {C\left( {6,1} \right)}_{{\rm{say}}\,\,D\,\,\left( {3{\rm{rd}}} \right)}\, \hfill \cr \\
\,{\rm{total}}\,\, = \,\,9 \cdot 8 \cdot 7\,\,{\rm{equiprobables}}\,\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,P\left( {{\rm{ii}}} \right) = {3 \over {14}}\)


\(? = {1 \over {84}} + {{3 \cdot 6} \over {14 \cdot 6}} = {{19} \over {84}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

The total number of ways to select 3 people from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

The number of ways 2 of the triplets are among the 3 awarded is 3C2 x 6C1 = 3 x 6 = 18.

The number of ways all 3 triplets are the 3 awarded is 3C3 x 6C0 = 1 x 1 = 1.

Therefore, the probability is (18 + 1)/84 = 19/84.

Answer: B

Hi Bunuel isn't the probability that at least two of the triplets will win a medal equal to none of the triplets win the medal? But the answer coming through ths way isnt correct.. 1- 5/21 = 16/21

At least two of the triplets will win a medal, means that only two or all three of the triplets win a medal. The opposite event would be that only one or none of the triplets win a medal.

Since the other solutions in this thread use counting techniques, let's use probability rules to solve the question

P(at least two triplets win a metal) = P(exactly two triplets win a middle OR exactly three triplets win a medal)
= P(exactly two win a middle) + P(exactly three win a medal)

Let's examine each probability separately....

P(exactly two triplets win a middle)
There are three ways for this to happen:
case i) a triplet places 1st, another triplet places 2nd, a non-triplet places 3rd
case ii) a triplet places 1st, a non-triplet places 2nd, a triplet places 3rd
case iii) a non-triplet places 1st, a triplet places 2nd, a non-triplet places 3rd

P(case i) = (3/9)(2/8)(6/7) = 1/14
P(case ii) = (3/9)(6/8)(2/7) = 1/14
P(case iii) = (6/9)(3/8)(2/7) = 1/14
P(exactly two triplets win a middle) = 1/14 + 1/14 + 1/14 = 3/14


P(exactly three triplets win a middle)
P(exactly three triplets win a middle) = P(a triplet places 1st, another triplet places 2nd, another triplet places 3rd)
= (3/9)(2/8)(1/7) = 1/84

We get:
P(at least two triplets win a metal) = P(exactly two win a middle) + P(exactly three win a medal)
= 3/14 + 1/84
= 18/84 + 1/84
= 19/84

Answer: B

Cheers,
Brent

Alternate approach:

Case 1: Exactly two triplets win medals
P(a triplet wins the first medal) \(= \frac{3}{9}\) (Of the 9 competitors, 3 are triplets)
P(a triplet wins the second medal) \(= \frac{2}{8}\) (Of the 8 remaining competitors, 2 are triplets)
P(a non-triplet wins the third medal) \(= \frac{6}{7}\) (Of the 7 remaining competitors, 6 are non-triplets)
To combine these probabilities, we multiply:
\(\frac{3}{9} * \frac{2}{8} * \frac{6}{7}\)
Since the non-triplet can be first, second, or third, we multiply by 3:
\(\frac{3}{9} * \frac{2}{8} * \frac{6}{7} * 3 = \frac{3}{14}\)

Case 2: All 3 triplets win a medal
P(a triplet wins the first medal) \(= \frac{3}{9}\) (Of the 9 competitors, 3 are triplets)
P(a triplet wins the second medal) \(= \frac{2}{8}\) (Of the 8 remaining competitors, 2 are triplets)
P(a triplet wins the third medal) \(= \frac{1}{7}\) (Of the 7 remaining competitors, 1 is a triplet)
To combine these probabilities, we multiply:
\(\frac{3}{9 }* \frac{2}{8} *\frac{ 1}{7} = \frac{1}{84}\)

Since a favorable outcome will be yielded by Case 1 or Case 2, we ADD the fractions:
\(\frac{3}{14} + \frac{1}{84} = \frac{19}{84}\)

Hi Bunuel,

I have a question:

Shouldn't the order in which the triplets win the medals matter? For instance, a case in which A wins Gold and B wins Silver is different from the case in which A wins Silver and B wins gold.

However, I do understand that this would mean 3! will be multiplied in each case as well as in the total number of cases (i.e., the denominator), which would result in them being cancelled out and hence, the answer wouldn't change.

But still, curious to understand from a process perspective.

Thanks again for your time!

here as u did by permutation u have to reduce it to combination ie

for first factor divide it by 2!/3! and second term by 3!/3! so it means only first term u r multiplying by 3. to get 18/84 + 1/84 gives 19/84

There are two ways we can answer this question.
1. Solve P(at least two win)
2. Solve 1 - (P(zero win) + P(exactly one wins))

1. Solve P(at least two win) = P(exactly two win) + P(exactly three win)
There are two scenarios: exactly two win and exactly three win the triathlon

Exactly two win the triathlon
Let X represent Adam, Bruce, or Charlie, and let Y represent the rest

P(XXY) (Two of the triplets win the 1st and 2nd place, while the either of the rest win the third place)
(3/9) x (2/8) x (6/7) = 1/14

P(XYX)
(3/9) x (6/8) x (2/7) = 1/14

P(YXX)
(6/9) x (3/8) x (2/7) = 1/14

P(exactly two win) = P(XXY) + P(XYX) + P(YXX) = 1/14 + 1/14 + 1/14 = 3/14

Note:
Why did I subtract the denominator by 1? Say, for example, the XXY scenario.

First, I need to choose either of the the triplets among nine people as the first winner. Therefore, the probability will be 3/9.
Second, I need to choose either of the two (the triplets) among the remaining eight people, since the first winner has been chosen. (2/8)
Lastly, I have to choose either of the six people as the third winner among the remaining seven people. (6/7)
Thus, the probability of XXY is (3/9) x (2/8) x (6/7) = 1/14

Next, there are three other scenarios: XXY, XYX, or YXX. You can count it just as I did, or you can just count 3!/2! = 3 to get the number of all scenarios (of exactly two win) and multiply it by 1/14 = 3/14

Exactly three win the triathlon
There is only one scenario, which is XXX

P(XXX) = (3/9) x (2/8) x (1/7) = 1/84

P(at least two win) = P(exactly two) + P(exactly three) = (3/14) + (1/84) = 19/84


2. Solve 1 - (P(exactly one) + P(exactly one)

Exactly no wins the triathlon
P(YYY)
(6/9) x (5/8) x (4/7) = 5/21

Exactly one wins the triathlon
P(XYY)
(3/9) x (6/8) x (5/7) = 5/28

P(YXY)
(6/9) x (3/8) x (5/7) = 5/28

P(YYX)
(6/9) x (5/8) x (3/7) = 5/28

P(exactly one wins) = P(XYY) + P(YXY) + P(YYX) = 5/28 + 5/28 + 5/28 = 15/28

Or you can count 3!/2! = 3 to arrive at the number of all possible scenarios (of exactly one win) and multiply it by 5/28 = 15/28

P(at least two win) = 1 - (P(exactly no wins) + P(exactly one win) = 1 - (5/21+ 15/28) = 19/84

Thank you!

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