Manhattan GMAT 700+ problem, June 26, 2006

[KevinFitzgerald]

Most of our students 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

[bdsam]

Originally Posted by KevinFitzgerald

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

My Take: B
Here goes my solution!

Now the number of ways 3 positions can be taken by 9 people is 9 * 8 * 7 [i.e. 9 P 3]

So, to find the peobability of atleast two positions going to the Triplets we need to find the probapility as:

Required Probability = Probability of 2 position going to triplets + Probability of all three position going to triplets

ie [for our refference]: P = P(2) + P(3)

Now P(2) = Triplets(1st & Second pos) + Triplets(Second & 3rd pos) + Triplets(1st & 3rd pos)

Therefire, P(2) =[(3 * 2 * 6)/ (9 * 8 * 7) + (6 * 3 * 2)/ (9 * 8 * 7) + (3 * 6 * 2)/ (9 * 8 * 7)]

=> p(2) = 108/(9 * 8 * 7)

Again, P(3) = (3 * 2 * 1) / (9 * 8 * 7)

So, P = 114/(9 * 8 * 7)
=> P = 19/84

Alternatively,

You can also get the same result by suptracting Probability of none of them getting Medal and probability of only one of them getting medal from 1

i.e: P = 1 - [P(0) + p(1)]

[ahmadazeem777]

B for me too.

Thankx
Azeem

[mals24]

Originally Posted by bdsam

My Take: B
Here goes my solution!

Now the number of ways 3 positions can be taken by 9 people is 9 * 8 * 7 [i.e. 9 P 3]

So, to find the peobability of atleast two positions going to the Triplets we need to find the probapility as:

Required Probability = Probability of 2 position going to triplets + Probability of all three position going to triplets

ie [for our refference]: P = P(2) + P(3)

Now P(2) = Triplets(1st & Second pos) + Triplets(Second & 3rd pos) + Triplets(1st & 3rd pos)

Therefire, P(2) =[(3 * 2 * 6)/ (9 * 8 * 7) + (6 * 3 * 2)/ (9 * 8 * 7) + (3 * 6 * 2)/ (9 * 8 * 7)]

=> p(2) = 108/(9 * 8 * 7)

Again, P(3) = (3 * 2 * 1) / (9 * 8 * 7)

So, P = 114/(9 * 8 * 7)
=> P = 19/84

Alternatively,

You can also get the same result by suptracting Probability of none of them getting Medal and probability of only one of them getting medal from 1

i.e: P = 1 - [P(0) + p(1)]

I have a doubt how come you've multiplied 3*2*6 to find the P2. should'nt it be 3*2*7.
If you take 3*2*7 the answer would be C 11/42