It is currently 23 Jun 2017, 22:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# GMAT Diagnostic Test Question 38

Author Message
Founder
Joined: 04 Dec 2002
Posts: 15143
Location: United States (WA)
GMAT 1: 750 Q49 V42
GMAT Diagnostic Test Question 38 [#permalink]

### Show Tags

07 Jun 2009, 01:02
Expert's post
20
This post was
BOOKMARKED
GMAT Diagnostic Test Question 38
Field: probability
Difficulty: 750

At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. $$\frac{1}{12}$$

B. $$\frac{5}{14}$$

C. $$\frac{4}{9}$$

D. $$\frac{1}{2}$$

E. $$\frac{2}{3}$$
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Last edited by bb on 29 Sep 2013, 21:47, edited 2 times in total.
Updated
CIO
Joined: 02 Oct 2007
Posts: 1218

### Show Tags

08 Jul 2009, 06:30
5
KUDOS
3
This post was
BOOKMARKED
Explanation:

In order to answer the question we need to find the overall number of outcomes and the number of favourable outcomes. The favourable outcome in this case is the one when a contestant tastes only 2 kinds of tea out of 3 kinds available. If there are 3 cups of every kind of tea, the number of favourable outcomes is calculated in the following way:

$$C_6^4 * 3 = 3 * \frac{6!}{4!*2!} = 45$$

We had to multiply by 3 because there are 3 ways the two kinds of tea could be selected from 3 available kinds.

The overall number of outcomes is equal to

$$C_9^4 = \frac{9!}{5!*4!} = \frac{9*8*7*6}{4*3*2} = 126$$

So, the probability can be found:

$$P = \frac{45}{126} = \frac{5}{14}$$
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 11 Dec 2008
Posts: 478
Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9

### Show Tags

20 Jul 2009, 08:46
dzyubam wrote:
Explanation:
 Rating:

In order to answer the question we need to find the overall number of outcomes and the number of favourable outcomes. The favourable outcome in this case is the one when a contestant tastes only 2 kinds of tea out of 3 kinds available. If there are 3 cups of every kind of tea, the number of favourable outcomes is calculated in the following way:

$$C_6^4 * 3 = 3 * \frac{6!}{4!*2!} = 45$$

We had to multiply by 3 because there are 3 ways the two kinds of tea could be selected from 3 available kinds.
$$P = \frac{45}{126} = \frac{5}{14}$$

I don't really understand how you got 45. I think you might be double counting since you are basically choosing 2 out of 6. However, that still leaves the possibility that the 3rd choice might make a complete set of 3... can you clarify why you are doing this again?

Basically, after computing the total number of combinations, I counted the number of "winning" combinations. If you are choosing 4 out of 9 from 3 sets of 3 and you don't want there to be a complete set, the only possibly combinations are either 3-1-0 or 2-2-0. If you choose 3 of 1 sample, you can either choose 3-0-1 or 3-1-0. Since there are 3 samples in all, that makes 3*2 = 6 combinations.

If choose 2 of one sample and 2 of another, then the only combinations are 2-2-0, 2-0-2, and 0-2-2.

The total number of combinations is then 9, so the probability is 9/126 = 1/14.

Can someone tell me if/where I went wrong?

Also, the question needs to be more clear. You need to state that its 9 marked cups of 3 samples each, otherwise you have to assume that the samples are equally divided.
Current Student
Joined: 13 Jul 2009
Posts: 145
Location: Barcelona
Schools: SSE
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

20 Jul 2009, 12:32
1
KUDOS
1
This post was
BOOKMARKED
Hello bipolarbear,
you are basically missing combinations:

3 0 1 x 3 = 3
3 1 0 x 3 = 3
0 1 3 x 3 = 3
1 0 3 x 3 = 3
0 3 1 x 3 = 3
1 3 0 x 3 = 3 (I multiply by 3 because in the sample with 1 cup you have 3 alternatives)

2 0 2 x 9 = 9
0 2 2 x 9 = 9
2 2 0 x 9 = 9 (you have 3 different 2-cup combination in each 2-cup sample, so 3x3)

TOTAL $$\frac{45}{126} = \frac{5}{14}$$
_________________

Performance: Gmat | Toefl
Contributions: The Idioms Test | All Geometry Formulas | Friendly Error Log | GMAT Analytics
MSc in Management: All you need to know | Student Lifestyle | Class Profiles

Founder
Joined: 04 Dec 2002
Posts: 15143
Location: United States (WA)
GMAT 1: 750 Q49 V42

### Show Tags

20 Jul 2009, 12:38
bipolarbear wrote:

Also, the question needs to be more clear. You need to state that its 9 marked cups of 3 samples each, otherwise you have to assume that the samples are equally divided.

Thank you!
Very good point - revising right now.
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Senior Manager
Joined: 11 Dec 2008
Posts: 478
Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

20 Jul 2009, 12:43
saruba wrote:

(you have 3 different 2-cup combination in each 2-cup sample, so 3x3)

TOTAL $$\frac{45}{126} = \frac{5}{14}$$

oh darn, you are clever. i stand corrected.
SVP
Joined: 29 Aug 2007
Posts: 2473
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

20 Jul 2009, 21:35
2
KUDOS
bb wrote:
GMAT Diagnostic Test Question 39
Field: probability
Difficulty: 750
 Rating:

At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. $$\frac{1}{12}$$
B. $$\frac{5}{14}$$
C. $$\frac{4}{9}$$
D. $$\frac{1}{2}$$
E. $$\frac{2}{3}$$

The highlighted part is not very clear however it is a basic combination problem.

1. 3 types of tea each has 3 cups totaling 9 cups.
2. Select 4 cups of tea of 2 different types out of 9 cups.

The number of ways 2 types of teas can be selected from 3 types of tea = 3c2 = 3 ways
The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways
The number of ways 4 cups of tea can be selected from 2 types of tea in 3 different ways = 3x15 = 45 ways
The number of ways 4 cups of tea can be selected from all 3 types of tea = 9c4 = 9x8x7x6x5!/(5!4!) = 146 ways

The prob that 4 cups of tea can be selected only from 2 types of tea = 45/126 = 15/42 ways

I do not see any complication with the OA and OE.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Intern
Joined: 30 May 2009
Posts: 3
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

23 Jul 2009, 16:10
11
KUDOS
1
This post was
BOOKMARKED
probability that a contestant does not taste all of the samples = 1 - probability that a contestant tastes all of the samples

now, lets calculate probability that a contestant tastes all of the samples.

there are 3 cases for this, as there are three samples i. e. 2,1,1(two cups of 1st sample, 1 cup of 2nd sample, 1 cup of 3rd sample), 1,2,1 and 1,1,2.

probability that a contestant tastes all of the samples = (3C2 * 3C1 * 3C1 + 3C1 * 3C2 * 3C1 + 3C1 * 3C1 * 3C2)/ 9C4
= (3*27)/(63*2)
= 9/14

therefore,
probability that a contestant does not taste all of the samples = 1 - 9/14
= 5/14
Manager
Joined: 08 Jul 2009
Posts: 172
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

29 Jul 2009, 12:45
1
KUDOS
vishalgupta wrote:
probability that a contestant does not taste all of the samples = 1 - probability that a contestant tastes all of the samples

now, lets calculate probability that a contestant tastes all of the samples.

there are 3 cases for this, as there are three samples i. e. 2,1,1(two cups of 1st sample, 1 cup of 2nd sample, 1 cup of 3rd sample), 1,2,1 and 1,1,2.

probability that a contestant tastes all of the samples = (3C2 * 3C1 * 3C1 + 3C1 * 3C2 * 3C1 + 3C1 * 3C1 * 3C2)/ 9C4
= (3*27)/(63*2)
= 9/14

therefore,
probability that a contestant does not taste all of the samples = 1 - 9/14
= 5/14

Great solution vishal
Intern
Joined: 27 Aug 2009
Posts: 1
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

27 Aug 2009, 22:16
19
KUDOS
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$
CIO
Joined: 02 Oct 2007
Posts: 1218
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

28 Aug 2009, 01:53
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 14 Dec 2009
Posts: 76
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

22 Dec 2009, 04:43
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

Great approach! +1
Intern
Joined: 25 Apr 2009
Posts: 10
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

25 Dec 2009, 13:07
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

very excellent approach +1
Intern
Joined: 02 Nov 2009
Posts: 10
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

17 Oct 2010, 12:24
Quote:
The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways

Hi,
can someone pls explain how this is 6c2, as there is no way to distinguish between the cups of the same sample.
The post of gmatclub math book (math-combinatorics-87345.html) clearly says that:

"Number of ways to pick 1 or more objects from n identical objects = n"

In this case, there are 3 ways a given cup can be picked up from one set of 3 cups and there are 2 ways to pick up that set of cups itself..i.e. there are 2 * 3 = 6 ways of picking up 4 cups.

Hence 3c2 * 6/9c4 = 1/7 should be the answer..?
Pls explain. Eagerly waiting.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 39622
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

19 Oct 2010, 15:23
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
gmatbestshot wrote:
Quote:
The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways

Hi,
can someone pls explain how this is 6c2, as there is no way to distinguish between the cups of the same sample.
The post of gmatclub math book (math-combinatorics-87345.html) clearly says that:

"Number of ways to pick 1 or more objects from n identical objects = n"

In this case, there are 3 ways a given cup can be picked up from one set of 3 cups and there are 2 ways to pick up that set of cups itself..i.e. there are 2 * 3 = 6 ways of picking up 4 cups.

Hence 3c2 * 6/9c4 = 1/7 should be the answer..?
Pls explain. Eagerly waiting.

Thanks

This question was posted in PS subforum as well, so below is my solutions from there. Hope these solutions will help to clear your doubts.

At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

# $$\frac{1}{12}$$
# $$\frac{5}{14}$$
# $$\frac{4}{9}$$
# $$\frac{1}{2}$$
# $$\frac{2}{3}$$

"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).

$$\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}$$.

$$C^2_3$$ - # of ways to choose which 2 samples will be tasted;
$$C^4_6$$ - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.

Another way:

Calculate the probability of opposite event and subtract this value from 1.

Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).

$$C^1_3$$ - # of ways to choose the sample which will provide with 2 cups;
$$C^2_3$$ - # of ways to chose these 2 cups from the chosen sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from second sample;
$$C^1_3$$ - # of ways to chose 1 cup out of 3 from third sample;
$$C^4_9$$ - total # of ways to choose 4 cups out of 9.

$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.

Hope it's clear.
_________________
Senior Manager
Joined: 19 Apr 2011
Posts: 277
Schools: Booth,NUS,St.Gallon
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

07 Feb 2012, 00:14
To calculate the number of favourable outcomes let us consider two cases .
1.three cups of a single sample and one cup from from the other two samples .
Number of ways =3*6c1=18.[ 3-->denotes number of ways in which three cups from a single sample can be selected ]

2.two cups each from any two samples .
Number of ways =3*3c2*3c2=27.[3-->denotes the number of ways of selecting 2 samples out of the 3 samples ]

Total number of favourable cases =18+27=45.

Total number of cases =9c4=126

Required propability=45/126=5/14.
_________________

+1 if you like my explanation .Thanks

Intern
Joined: 13 Jul 2012
Posts: 4
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

13 Jul 2012, 21:04

Firstly, we have to eliminate the probability of taking all the samples after 4 tastes. There are 3 ways to do that.

1/ He took 3 different samples in the 3 first tastes, so we don't care whatever the last time he took.
Probability of this way = 1*6/8*3/7*1=9/28

2/ The 2nd time he took the same sample with the 1st one and the 3rd and 4th time he took the 2 others.
Probability = 1*2/8*6/7*3/6=3/28

3/ He took 2 different samples in the 2 first time; the 3rd time he took the same sample with either the 1st or the 2nd one, and the 4th time he took the other sample.
Probability = 1*6/8*4/7*3/6=6/28

So the answer = 1-(9/28+3/28+6/28) = 10/28 = 5/14
Director
Joined: 28 Jun 2011
Posts: 889
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

22 Aug 2013, 22:47
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

Easiest approach of all.
_________________
Intern
Joined: 04 Oct 2013
Posts: 3
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

07 Oct 2013, 15:26
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

Sorry, I can't follow this part, can anyone tell me where the multiplication by 3 comes from?

Quote:
since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$
Math Expert
Joined: 02 Sep 2009
Posts: 39622
Re: GMAT Diagnostic Test Question 39 [#permalink]

### Show Tags

08 Oct 2013, 00:50
hsarci wrote:
saurabhmukim wrote:
Pls suggest if this approach is acceptable...

There are total 9 cups... In order to NOT to taste from one type, the contestant has to choose from remaining 6..

i.e. $$\frac{6}{9}$$.. similarly for the next cup, it choice would be $$\frac{5}{8}$$ and so on....

4 cups implies $$\frac{6*5*4*3}{9*8*7*6}$$.... which is $$\frac{5}{42}$$..

since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

Sorry, I can't follow this part, can anyone tell me where the multiplication by 3 comes from?

Quote:
since there are three types of tea available $$\frac{3*5}{42}$$ which equals $$\frac{5}{14}$$

The following post might help: gmat-diagnostic-test-question-79371.html#p803095
_________________
Re: GMAT Diagnostic Test Question 39   [#permalink] 08 Oct 2013, 00:50
Similar topics Replies Last post
Similar
Topics:
42 GMAT Diagnostic Test Question 5 26 20 Jun 2014, 03:49
16 GMAT Diagnostic Test Question 4 37 09 Apr 2014, 06:43
25 GMAT Diagnostic Test Question 3 33 22 Oct 2013, 23:26
9 GMAT Diagnostic Test Question 2 26 04 Oct 2013, 03:27
36 GMAT Diagnostic Test Question 1 30 14 Apr 2014, 02:53
Display posts from previous: Sort by

# GMAT Diagnostic Test Question 38

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.