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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
06 Mar 2013, 01:32

Expert's post

tosattam wrote:

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?

A blind taste simply means that a contestant doesn't know which samples he/she is offered to taste. _________________

Re: Diagnostic Question [#permalink]
09 Mar 2013, 21:37

1

This post received KUDOS

Hi, Let me try.. mainhoon, you are going with the logic that you select 1 cup from each sample in 3c1 x 3c1 x 3c1 ways and 1 remaining cup from 6 cups in 6c1 ways rite! I guess the flaw with your approach is that this 6c1 is the probablity of selecting 1 item from 6 DIFFERENT items (r items from n different items is nCr) Im this case the 6 items left are not different! They are XX,YY,ZZ types right! so you just cant use the 6c1 fomula! This is the formula : The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of x^r in the expansion

hope it helps!

mainhoon wrote:

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?

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
29 Jan 2014, 10:59

If we employ the counting method applied in the question http://gmatclub.com/forum/tough-p-n-c-92675.html then Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another i.e. AAAB The number of ways this can happen is 3C2*4!/3! = 12 3C2 = Ways to select any 2 samples out of the 3 choices 4!/3! = # of ways AAAB can be rearranged

or 2. Contestant tastes 2 cups of a sample and 2 of another i.e. AABB The number of ways this could happen is 3C2*4!/(2!*2!) = 18 3C2 = Ways to select any 2 samples out of the 3 choices 4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126 = 5/21

Is this approach not correct? Where am I missing ?

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
29 Jan 2014, 11:22

Great explanations above. A rather headache-inducing question, I admit I had to resolve to the following guessing pattern after I passed the 1:36 mark and admitted defeat.

Economist wrote:

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. \(\frac{1}{12}\) Too low B. \(\frac{5}{14}\) Looks reasonable C. \(\frac{4}{9}\) Impossible, cannot be that simple D. \(\frac{1}{2}\) Too high E. \(\frac{2}{3}\) Too high

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
30 Jan 2014, 01:50

Quote:

If we employ the counting method applied in the question tough-p-n-c-92675.html then Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another i.e. AAAB The number of ways this can happen is 3C2*4!/3! = 12 3C2 = Ways to select any 2 samples out of the 3 choices 4!/3! = # of ways AAAB can be rearranged

or 2. Contestant tastes 2 cups of a sample and 2 of another i.e. AABB The number of ways this could happen is 3C2*4!/(2!*2!) = 18 3C2 = Ways to select any 2 samples out of the 3 choices 4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126 = 5/21

Is this approach not correct? Where am I missing ?

I have come up with another approach which gives a different result again.

Assuming the that the contestant does not taste the 3rd sample there are 2 scenarios possible

AABB and AAAB

Counting one at a time. #of ways AABB can occur = 3C2*3C2*3C2 = 27 here, 3C2 = ways to select the 2 samples from 3 3C2 = ways to select the 2 cups from 3 cups of first sample 3C2 = ways to select 2 cups from 3 cups of second sample

# of ways AAAB can occur = 3C2*3C3*3C1 = 9 3C2 = ways to select the 2 samples from 3 3C3 = ways to select 3 cups from the 3 cups of first sample 3C1 = ways to select 1 cup from the 3 cups of second sample

Hence the total number of ways in which the contestant cannot taste any cup from third sample = 9 +27 = 36

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 36/126 = 4/14

Even after hours and hours of going through various material on Probablity and combinations I am struggling to identify a uniform approach that can be applied across problems. How would one know which approach to apply for which problem during the exams. Is there any way we can determine that?

At a blind taste competition a contestant is offered 3 cups [#permalink]
28 Oct 2014, 22:39

# of ways of choosing 4 cups from 9 = 9c4 = 126 # of ways of choosing 4 cups such that one gets only 2 varieties of tea -

1st, 2nd and 3rd cup to be from same type = 3*2*1 per type * 3 (for 3 different types) = 27 4th cup to be from a different type (can't be from same type as none of 1 type are left) = 3*3 per type * 2 (for 2 different types) = 18

Total = 1st, 2nd, 3rd & 4th cup = 27+18 = 45

Prob (Tastes only 2 varieties) = 45/126 = 5/14 - B.

At a blind taste competition a contestant is offered 3 cups [#permalink]
12 Jan 2015, 07:47

I solved it in a way that I think is more direct: I get the probability that each cup is not a specific type of tea (type A): 1st cup, Probability it is not type A = 3/9 2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A 3rd cup, Probability it is not type A = 3/7 4th cup, Probability it is not type A = 3/6 We do the same for types B and C (multiply by 3) So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
24 Mar 2015, 01:45

Hi Is the following approach not correct..?? where am wrong????

C^1_3 - # of ways to chose 1 cup out of 3 from 1st sample; C^1_3 - # of ways to chose 1 cup out of 3 from 2nd sample; C^1_3 - # of ways to chose 1 cup out of 3 from third sample; C^1_6- # of ways to chose 1 cup out of 6 remaining cups;

C^4_9- # of ways to chose 4 cup out of 9 cups;

Probability = [C^1_3 x C^1_3 x C^1_3 x C^1_6] / C^4_9 = 162/126 =9/7

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
07 Sep 2015, 02:25

Hi,

Can someone help me understand the question? I am having great trouble figuring out what exactly is happening here, let alone what's asked.

"3 cups of each of the 3 samples of tea" - does this mean that there are 3 types of tea, say Aasam Tea, Green Tea and Ginger Tea and a contestant is given 1 cup of each?

"in a random arrangement of 9 marked cups" - what does this mean? Are there 9 cups, each with a different "type" of tea, or are there just 9 cups that may/may not have tea? The three cups offered above - these come from these nine cups?

"each contestant tastes 4 different cups of tea" - each contestant as against "a" contestant? So what's happening here? Each tastes 4 cups, 3 of which he has been given as part of the first point above (all three of a different flavour + one he picks up on his own?)

"contestant does not taste all of the samples" - so probability that he does not taste each of aasam,green and ginger? As in, a contestant tastes either zero or one or two? What's the difference here between a contestant and each contestant?

I am completely confused! Can someone help explain what's happening with a diagram? Thanks in advance! _________________

If you like this post, be kind and help me with Kudos!

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
18 Oct 2015, 23:04

Bunuel wrote:

Economist wrote:

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.

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
01 Nov 2015, 16:07

dimitri92 wrote:

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

Hi! Can some please explain the note regarding "there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1."? I didn't understand why we need to multiple by 2.

Separately, why is my approach below wrong?

P= 1-(9*6*3*6)/(9 choose 4) is obviously wrong, but why? The denominator I understand very clearly - number of ways to choose 4 from 9 choices. The numerator is based on the reasoning that the first cup can be from any of the 9 choices, so 9 ways. The second cup has to be from any of the 6 cups (the two samples different from the first cup). The third cup has to be from any of the 3 cups (the third sample not choose by cup 1 or 2). And the last cup can be from any of the remaining 6 cups.

Thanks so much! _________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
23 Nov 2015, 18:06

dimitri92 wrote:

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

Hi Bunnuel, Is this a correct solution? I am having a difficult time understanding why did he multiply by two. I would expect that through this method, you don't need to multiply to account for combinations since the probabilities for each pick already account for that. Thanks.

Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
01 Dec 2015, 21:45

happyface101 wrote:

dimitri92 wrote:

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

Hi! Can some please explain the note regarding "there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1."? I didn't understand why we need to multiple by 2.

Separately, why is my approach below wrong?

P= 1-(9*6*3*6)/(9 choose 4) is obviously wrong, but why? The denominator I understand very clearly - number of ways to choose 4 from 9 choices. The numerator is based on the reasoning that the first cup can be from any of the 9 choices, so 9 ways. The second cup has to be from any of the 6 cups (the two samples different from the first cup). The third cup has to be from any of the 3 cups (the third sample not choose by cup 1 or 2). And the last cup can be from any of the remaining 6 cups.

Thanks so much!

BUMP my question - please help! _________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

gmatclubot

Re: At a blind taste competition a contestant is offered 3 cups
[#permalink]
01 Dec 2015, 21:45

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