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Combinatorics

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by rishab1988 » Mon Dec 06, 2010 6:48 am
A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 90 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90 ). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?


(A) 3/16000

(B) 1/8100

(C) 3/1600

(D) 1/90

(E) 2/45
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by shovan85 » Mon Dec 06, 2010 7:10 am
rishab1988 wrote:A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 90 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90 ). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?


(A) 3/16000

(B) 1/8100

(C) 3/1600

(D) 1/90

(E) 2/45
Probability of choosing a cello = 90/800
Probability of choosing that particular viola from the tree= 1/600

probability that the two instruments are made with wood from the same tree = 90/800 * 1/600 = 3/16000

IMO A
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by Rahul@gurome » Mon Dec 06, 2010 7:18 am
Number of ways to select a cello-viola pair from 800 cellos and 600 violas = (Number of ways to select one cello from 800)*(Number of ways to select a viola from 600) = (800)*(600)

Number of ways to select a cello-viola pair from 800 cellos and 600 violas such that they are from same pair = (Number of ways to select one cello from 90)*(Number of ways to select the viola of the same pair from 90) = (90)*(1)

Required probability = 90/(600*800) = 3/16000

The correct answer is A.
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by miguelmickelberg » Mon Dec 06, 2010 7:50 am
Got this one right. Does anyone know the difficulty level?
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by rishab1988 » Mon Dec 06, 2010 7:54 am
It is above 700+.You don't get combinatorics questions unless you are scoring 48+ on quant.

Also you can't judge your performance on quant just by taking 1 question as a metric.You got to perform consistently and under timed conditions!

Further,if you are so good at combinatorics,work hard on the other questions so that you can get such question :)
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by miguelmickelberg » Mon Dec 06, 2010 8:04 am
Rishab, you are definitely right. I answered it in 1:10, timing it with the timer at the top. Anyway, this doesn't mean much, I just wanted to have an idea on how much I need to improve in combinatorics. I still have a long way on inequalities, word problems, etc. But at least I can get a good number of combinatorics questions right.
rishab1988 wrote:It is above 700+.You don't get combinatorics questions unless you are scoring 48+ on quant.

Also you can't judge your performance on quant just by taking 1 question as a metric.You got to perform consistently and under timed conditions!

Further,if you are so good at combinatorics,work hard on the other questions so that you can get such question :)
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by goyalsau » Mon Dec 06, 2010 9:02 am
miguelmickelberg wrote:Rishab, you are definitely right. I answered it in 1:10, timing it with the timer at the top. Anyway, this doesn't mean much, I just wanted to have an idea on how much I need to improve in combinatorics. I still have a long way on inequalities, word problems, etc. But at least I can get a good number of combinatorics questions right.
rishab1988 wrote:It is above 700+.You don't get combinatorics questions unless you are scoring 48+ on quant.

Also you can't judge your performance on quant just by taking 1 question as a metric.You got to perform consistently and under timed conditions!

Further,if you are so good at combinatorics,work hard on the other questions so that you can get such question :)
It's Nice to see that people are really good in Combinatorics, I just don't know when i will able to answer them correctly, :cry: :cry:
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by rishab1988 » Mon Dec 06, 2010 9:25 am
goyalsau wrote:
miguelmickelberg wrote:Rishab, you are definitely right. I answered it in 1:10, timing it with the timer at the top. Anyway, this doesn't mean much, I just wanted to have an idea on how much I need to improve in combinatorics. I still have a long way on inequalities, word problems, etc. But at least I can get a good number of combinatorics questions right.
rishab1988 wrote:It is above 700+.You don't get combinatorics questions unless you are scoring 48+ on quant.

Also you can't judge your performance on quant just by taking 1 question as a metric.You got to perform consistently and under timed conditions!

Further,if you are so good at combinatorics,work hard on the other questions so that you can get such question :)
It's Nice to see that people are really good in Combinatorics, I just don't know when i will able to answer them correctly, :cry: :cry:
You don't need to be a champ at Combinatorics.In my 2 previous attempts,all I got were basic probability questions.This is one area in GMAT [if you reach!] where you can milk some easy points.
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by aditya_velma » Wed Dec 08, 2010 4:30 am
I have a query here, when I say (90/800)*(1/600), why shouldn't I add (90/600)*(1/800) to this?
I am using multiplication here, so why shouldn't I consider (cello first AND viola pair) OR (viola first AND cello pair)?

Thanks :)
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by Night reader » Wed Dec 08, 2010 3:26 pm
rishab1988 wrote:A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 90 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90 ). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?


(A) 3/16000

(B) 1/8100

(C) 3/1600

(D) 1/90

(E) 2/45
took around 1 min =>

the probability of drawing cello made of tree is 90/800
the probability of drawing viola exactly the same tree (unique, only one exists in population) is 1/600

combining two probabilities 9/80 * 1/600 = 3/1600
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by saxenankit » Sun May 01, 2011 5:47 am
aditya_velma wrote:I have a query here, when I say (90/800)*(1/600), why shouldn't I add (90/600)*(1/800) to this?
I am using multiplication here, so why shouldn't I consider (cello first AND viola pair) OR (viola first AND cello pair)?

Thanks :)
Dear experts,

Can any one of you please answer this ?

Thanks,
Ankit
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by chendawg » Wed Jun 15, 2011 5:57 pm
I'm having the same issues as the last poster; I don't get why we don't multiply by 2 @ the end since there are 2 ways to get the pair.

The solution to the problem above seems to me would be the same as(but obviously not) calculating the probability of getting a King, queen, or a Jack in three consecutive pulls (4/52*4/51*4/50*6); we would need to multiply by 6 because of the 6 different ways to pull a King, Queen, or a Jack.

Is the above problem kinda the same as the probability to get a pocket pair of Kings in a deck of cards (4/52*3/51)? I understand we don't need to multiply by two, because it's the same way each time.
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by lunarpower » Thu Jun 16, 2011 3:15 am
chendawg wrote:I'm having the same issues as the last poster; I don't get why we don't multiply by 2 @ the end since there are 2 ways to get the pair.

The solution to the problem above seems to me would be the same as(but obviously not) calculating the probability of getting a King, queen, or a Jack in three consecutive pulls (4/52*4/51*4/50*6); we would need to multiply by 6 because of the 6 different ways to pull a King, Queen, or a Jack.

Is the above problem kinda the same as the probability to get a pocket pair of Kings in a deck of cards (4/52*3/51)? I understand we don't need to multiply by two, because it's the same way each time.
if you are SELECTING THINGS FROM THE SAME POOL, *then* you must consider the different orders in which things may be selected.
if you are SELECTING THINGS FROM DIFFERENT POOLS, then you don't.

as proof, consider a stupid example: let's say that i have only 1 shirt, which is black, and only 1 pair of pants, which is white.
if i pick 1 shirt and 1 pair of pants, what's the probability that i get a black shirt and white pants?
(obviously the answer should be 1, since this is the only possible outcome.)
in this case, if you try what you're doing here, then you'll think the probability is (1)(1) + (1)(1) = 2, so something is definitely wrong with that.
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by GMATGuruNY » Thu Jun 16, 2011 3:38 am
rishab1988 wrote:A certain music store stocks 800 cellos and 600 violas. Of these instruments, there are 90 cello-viola pairs, such that a cello and a viola were both made with wood from the same tree (each tree can make at most one viola and one cello, so there are no pairs other than these 90 ). If one viola and one cello are chosen at random, what is the probability that the two instruments are made with wood from the same tree?


(A) 3/16000

(B) 1/8100

(C) 3/1600

(D) 1/90

(E) 2/45
I think of this as a "bucket" problem.
We have a bucket of cellos and a bucket of violas.

To combine from multiple buckets:

1. Determine how many choices can be made from each bucket.
2. Multiply the results.

Bucket of cellos = 800 choices.
Bucket of violas = 600 choices.

Total possible pairs = 800*600 = 480,000.

Of these, there are 90 good pairs.

P(good pair) = 90/480000 = 3/16000.

The correct answer is A.
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by chendawg » Thu Jun 16, 2011 9:10 am
lunarpower wrote: if you are SELECTING THINGS FROM THE SAME POOL, *then* you must consider the different orders in which things may be selected.
if you are SELECTING THINGS FROM DIFFERENT POOLS, then you don't.
So with this in mind, am I doing these scenarios and applying this principle correctly?

If I want to pull a King and Queen for hole cards (suits do not matter, nor does order), then the probability would be 8/663(4/52*4/51*2). I would be multiplying by two since the cards are from the same pool.

If I want to pull a King and Queen(suits don't matter, nor order), with a pull each from TWO standard decks, then it would be 4/169(8/52*8/52). I'm not multiplying by two this time because they are from pulls from separate decks each time.

If I want to pull a King and Queen(suits don't matter, nor order), with a pull from TWO standard decks, but this time we combine the decks and pull the cards, then it would be 60/1339(16/104*15/103*2). I would be multiplying by 2 this time since it's from the same pool.
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