Problem Solving

Most of our students Manhattan GMAT are trying to break the 700+ barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level Problem Solving problem (I'll post the solution next Monday).

Question:
Triplets Adam, Bruce, and Charlie enter a triathlon. If there are 9 competitors in the triathlon, and medals are awarded for first, second, and third place, 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

In my opinion : B.

==> (3C2*6C1) / (9C3) + (3C3) / (9C3)

Solving this: (18 + 1) / 84 = 19/84..

Answer
For probability, we always want to find the number of ways the requested event could happen and divide it by the total number of ways that any event could happen.

For this complicated problem, it is easiest to use combinatorics to find our two values. First, we find the total number of outcomes for the triathlon. There are 9 competitors; three will win medals and six will not. We can use the Combinatorics Grid, a counting method that allows us to determine the number of combinations without writing out every possible combination.

A B C D E F G H I
Y Y Y N N N N N N

Out of our 9 total places, the first three, A, B, and C, win medals, so we label these with a "Y." The final six places (D, E, F, G, H, and I) do not win medals, so we label these with an "N." We translate this into math: 9! / 3!6! = 84. So our total possible number of combinations is 84. (Remember that ! means factorial; for example, 6! = 6 × 5 × 4 × 3 × 2 × 1.)

Note that although the problem seemed to make a point of differentiating the first, second, and third places, our question asks only whether the brothers will medal, not which place they will win. This is why we don't need to worry about labeling first, second, and third place distinctly.

Now, we need to determine the number of instances when at least two brothers win a medal. Practically speaking, this means we want to add the number of instances two brothers win to the number of instances three brothers win.

Let's start with all three brothers winning medals, where B represents a brother.

A B C D E F G H I
B B B N N N N N N

Since all the brothers win medals, we can ignore the part of the counting grid that includes those who don't win medals. We have 3! / 3! = 1. That is, there is only one instance when all three brothers win medals.

Next, let's calculate the instances when exactly two brothers win medals.

A B C D E F G H I
B B Y B N N N N N

Since brothers both win and don't win medals in this scenario, we need to consider both sides of the grid (i.e. the ABC side and the DEFGHI side). First, for the three who win medals, we have 3! / 2! = 3. For the six who don't win medals, we have 6! / 5! = 6. We multiply these two numbers to get our total number: 3 × 6 = 18.

Another way to consider the instances of at least two brothers medaling would be to think of simple combinations with restrictions.

If you are choosing 3 people out of 9 to be winners, how many different ways are there to chose a specific set of 3 from the 9 (i.e. all the brothers)? Just one. Therefore, there is only one scenario of all three brothers medaling.

If you are choosing 3 people out of 9 to be winners, if 2 specific people of the 9 have to be a member of the winning group, how many possible groups are there? It is best to think of this as a problem of choosing 1 out of 7 (2 must be chosen). Choosing 1 out of 7 can be represented as 7! / 1!6! = 7. However, if 1 of the remaining 7 can not be a member of this group (in this case the 3rd brother) there are actually only 6 such scenarios. Since there are 3 different sets of exactly two brothers (B1B2, B1B3, B2B3), we would have to multiply this 6 by 3 to get 18 scenarios of only two brothers medaling.

The brothers win at least two medals in 18 + 1 = 19 circumstances. Our total number of circumstances is 84, so our probability is 19 / 84.

The correct answer is B.

(6C3 + 6C2*3C1)/9C3 = 65/84

so 1-p = 19/84 is the result

3 winners can be chosen from the group of 9 competitors in 9C3 ways = 84 ways

Atleast 2 twins have to win a medal. There are 2 scenarios

Scenario : When exactly 2 twins win a medal each

= 3C2 * 6C1 = 3*6 = 18 ways

Scenario II : When all three win a medal. There is just one way that this happens

= 3C3 = 1 way

Hence Probability = 18 + 1 = 19/84

Answer is B

Cybermusings wrote:3 winners can be chosen from the group of 9 competitors in 9C3 ways = 84 ways

Atleast 2 twins have to win a medal. There are 2 scenarios

Scenario : When exactly 2 twins win a medal each

= 3C2 * 6C1 = 3*6 = 18 ways

Scenario II : When all three win a medal. There is just one way that this happens

= 3C3 = 1 way

Hence Probability = 18 + 1 = 19/84

Answer is B

hey CM good job pullin out these old but good questions ... keep up the good work ...

To make this easier to solve (i.e at least two of the triplets who will win a medal). It can be see that,
(i) No of ways of choosing 2 out of the triplets = $$^{3_C}2\ =3$$
(ii) No of ways of choosing 3 out of the triplets = $$^{3_C}3\ =1$$
Also
(iii) No of ways of choosing 1 out of the other 6= $$^{6_C}1\ =6$$
(iv) making the awarded medals of 2 of the triplets and 1 out of the other , it can be calculated as $$^{3_C}2\ \cdot6C_1=18$$
(v) now total number of ways of selecting 3 of the medals awarded of the total 9= $$9C_3=84$$
from (ii) and (iv)
(vi) No of ways of choosing at least two = 18+1=19

from (vi) and (iv)
choosing at least two out of the total number of selecting 3 of the metals awarded 19/84