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At a blind taste competition a contestant is offered 3 cups [#permalink]
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Updated on: 06 Mar 2013, 02:30
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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. \(\frac{2}{3}\)
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Originally posted by Economist on 14 Nov 2009, 07:34.
Last edited by Bunuel on 06 Mar 2013, 02:30, edited 2 times in total.
Edited the question and added the OA



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Re: Prob gclub diagonostics [#permalink]
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14 Nov 2009, 08:06
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?
# \(\frac{1}{12}\) # \(\frac{5}{14}\) # \(\frac{4}{9}\) # \(\frac{1}{2}\) # \(\frac{2}{3}\) And the good one again. +1 to Economist. "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 (211). \(C^1_3\)  # 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. \(P=1\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1\frac{9}{14}=\frac{5}{14}\). Answer: B. Hope it's clear.
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Re: Prob gclub diagonostics [#permalink]
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15 Nov 2009, 04:16
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
Therefore,
\(P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}\)
\(P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}\)
\(P = \frac{45}{{9*2*7}}\)
\(P = \frac{5}{14}\)



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Diagnostic Question [#permalink]
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15 Aug 2010, 17:17
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... Ok so where did my brain get fried?
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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 18:43
It should be 1(3*3C2*3C1*3C1/9C4) = 19/14 = 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.
Hope it helps.



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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 18:50
Can you give the breakdown as to how you arrived at (3*3C2*3C1*3C1). Thanks My thinking was simply that  < select 1 of 3  < select 1 of 3  < select 1 of 3 That is a total of 27 ways. I can see that these are not unique combinations, but how do I remove the duplicates?
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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 19:03
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.
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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 19:16
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?
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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 19:53
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.



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Re: Diagnostic Question [#permalink]
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15 Aug 2010, 20:06
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?
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Re: Prob gclub diagonostics [#permalink]
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10 Nov 2010, 01:11
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...
+ 1 to Bunuel.....



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Re: Prob gclub diagonostics [#permalink]
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06 Feb 2011, 08:18
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 = \(1  \frac{9}{14} = \frac{5}{14}\)
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Re: Prob gclub diagonostics [#permalink]
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04 Aug 2011, 00:47
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?



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Re: Prob gclub diagonostics [#permalink]
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26 Aug 2011, 16:59
perhaps bunuel can chime in / elaborate on dimitri92's answer?
using the 9/9 * 6/8 * 2 * 3/7 * 6/6
this approach always confuses me.



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Re: Prob gclub diagonostics [#permalink]
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29 Aug 2011, 18:35
pinchharmonic wrote: perhaps bunuel can chime in / elaborate on dimitri92's answer?
using the 9/9 * 6/8 * 2 * 3/7 * 6/6
this approach always confuses me. bump on this? i'm curious about the 6/8 and the factor of 2. 9/9 i understand, but 6/8 should give the probability of either two types, then why is there another factor of 2?



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Re: Prob gclub diagonostics [#permalink]
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22 Dec 2011, 23:41
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?
# \(\frac{1}{12}\) # \(\frac{5}{14}\) # \(\frac{4}{9}\) # \(\frac{1}{2}\) # \(\frac{2}{3}\) And the good one again. +1 to Economist. "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 (211). \(C^1_3\)  # 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. \(P=1\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1\frac{9}{14}=\frac{5}{14}\). Answer: B. Hope it's clear. 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 \(\frac{6}{7}\)?



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



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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06 Mar 2013, 02:32



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Re: Diagnostic Question [#permalink]
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09 Mar 2013, 22:37
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?



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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29 Jan 2014, 11:59
If we employ the counting method applied in the question http://gmatclub.com/forum/toughpnc92675.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
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