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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

14 Nov 2009, 07:34

9

This post received KUDOS

25

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

61% (02:56) correct
39% (01:56) wrong based on 273 sessions

HideShow timer Statistics

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. \(\frac{2}{3}\)

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?

"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.

Answer: B.

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.

I got B: 5/14 as well, although using a different approach.

Number of ways to choose 4 cups to drink from: 9C4

Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.

Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2

Three cases:

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1 b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2 c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3

I thought the answer to this question can be obtained thus: 1 - P(all samples are tasted)

P(all samples tasted) = I choose one from each of the 3 samples = 3C1 x 3C1x 3C1 x 6C1 as there will be 6 cups left from which I can get the 4th cup = 27 x6 = 162 makes no sense as there are only 9C4 combinations = 126.

S1 A B C S2 D E F S3 G H I

Select 1 from S1 in 3 ways, same with S2, same with S3 = a total of 27 ways I have 6 cups left, can select 1 in 6 ways

So total of 27 x 6 = 162 ways. It seems that this should be 81, I am double counting somewhere. Only 81 gives me the right answer...

It should be 1-(3*3C2*3C1*3C1/9C4) = 1-9/14 = 5/14

Now why is your approach wrong ? E.g: in how many ways 2 objects can be selected from a set of 3. We know it is 3C2 As per the approach you took above it should be equal to; 3C1*2C1 != 3C2.

We want to select 2 cups of tea which are of similar type and the rest 2 cups which each are of different type ; hence the terms 3C2 * 3C1* 3C1 Now you should multiply it by 3 because ; out of the 3 tea types , you may select the tea type which is in 2 cups, in 3 ways.

Maybe I dont understand the question. But I was approaching it as 1 - p(he tastes all samples), so why limit to just 2 cups and then make it so that the other 2 are each of a different type?

S1 A B C S2 D E F S3 G H I

So why can't be go ADGF - select one from S1, one from S2, one from S3 and then have 1 of the remaining 6? You are doing 3c2 3c1 3c1, so that is picking 2 from one row and one each from the other 2 rows such as ABDG. Perhaps if you can explain how to get an accurate count of combinations, when I select 1 from each row, make it 3 cups and then pick the 4th one of the remaining 6? _________________

I am not sure if it could be calculated that way. Lets make it even simpler S1 A B C

In how many ways can you select 2 out of these 3 ? 3C2, right ? If however; if I first select 1 of them say in 3C1 ways; there are 2 still left. Now, if I again select 1 out of the remaining two, in 2C1 ways, the total no. of combination 3C1*2C1, which is not equal to 3C2.

Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this? _________________

I got B: 5/14 as well, although using a different approach.

Number of ways to choose 4 cups to drink from: 9C4

Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.

Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2

Three cases:

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1 b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2 c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3

I don't follow this answer. (a) and (c) show that he ends up tasting all the cups of the given sample, we are not supposed to allow this to happen? _________________

I was all over the place in trying to solve this...neways Thanks Bunuel..that was a magnificient and clear solution..but my only concern is on the actual GMAT..is this question easy enuf to be answered in under 2 mins...???or rather...it looks like a difficulty level 800+ to me...

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors. the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\) the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\) the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\) the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Can anyone tell me where I went wrong? I understand you can derive the answer from the solution above, but why does my method below give me a wrong answer?

I used 1 - (prob of drinking all three teas) Total combinations = 9C4 = 126 prob of drinking all threes = 1 of each tea + 1 of any tea. Using A, B, & C to represent the teas: Combinations where all three teas are represented (2 of one tea and one of each of the other tea) = A, A, B, C A, B, B, C A, B, C, C

Combinations of rearranging A,A,B,C = 4!/2! = 12 Combinations of rearranging A,B,B,C = 4!/2! = 12 Combinations of rearranging A,B,C,C = 4!/2! = 12

Total combinations of drinking each tea = 36 Probability of not drinking all tea = 1 - (36/126) = 90/126 = 5/7

I can't figure out where the logical error or where I double counted or ignored possibilities come from. Can anyone help?

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?

"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.

Answer: B.

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.

Thank you. If the question was about the probability that a contestant does not taste all the 3 samples of a cup. (That's what I wrongly understood from this question first). Then the answer would be \(\frac{6}{7}\)?

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

Show Tags

05 Mar 2013, 09:23

Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth. That gives a complete different perspective to the problem. Where am I going wrong here?

gmatclubot

Re: At a blind taste competition a contestant is offered 3 cups
[#permalink]
05 Mar 2013, 09:23

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...

Ghibli studio’s Princess Mononoke was my first exposure to Japan. I saw it at a sleepover with a neighborhood friend after playing some video games and I was...