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At a blind taste competition a contestant is offered 3 cups of each of
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Updated on: 06 Feb 2019, 23:35
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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 Feb 2019, 23:35, edited 3 times in total.
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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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: At a blind taste competition a contestant is offered 3 cups of each of
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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: At a blind taste competition a contestant is offered 3 cups of each of
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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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Re: At a blind taste competition a contestant is offered 3 cups of each of
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18 Mar 2016, 00:58
The probability of tasting all 3 cups is One cup from pool A (\(\frac{3}{9}\)) x One cup from pool B (\(\frac{3}{8}\)) x One cup from pool C (\(\frac{3}{7}\)) Since the contestant has to taste 4 cups, one possibility is picking second cup from pool A, so the probability is \(\frac{2}{6}\) The probability of tasting in order of ABCA is \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6}\) ..... (1) In how many ways can the contestant taste ABCA cups = \(\frac{4!}{2!}\) = 4 x 3 ....... (2) The combinations ABCB & ABCC are also possible (along with ABCA) = 3 combinations ..... (3) Since the choice of "a specific cup" from "a pool" doesn't matter here, we need NOT account for \(A_1A_2A_3, A_1A_3A_2,\) ... etc. combinations So Probability of tasting all three cups is = (1) * (2) * (3) = \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6} * 4 *3 *3\) =\(\frac{9}{14}\) Hence, Probability of NOT tasting all three cups = \(1 \frac{9}{14} = \frac{5}{14}\)
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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12 Jan 2015, 08:47
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
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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17 Jun 2019, 06:13
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 = \(1  \frac{9}{14} = \frac{5}{14}\)
I am having issues with the above, especially this line there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1. in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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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: At a blind taste competition a contestant is offered 3 cups of each of
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25 Sep 2017, 11:15
DanceWithFire wrote: cledgard wrote: 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 1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9 Interesting way to approach the problem, though. You are right, I made a mistake when I wrote it no the forum. The correct approach is the following: I get the probability that each cup is not a specific type of tea (type A): 1st cup, Probability it is not type A = 6/9 2nd cup, Probability it is not type A = 5/8, since we only have 8 cups left, and 3 of them are type A 3rd cup, Probability it is not type A =4/7 4th cup, Probability it is not type A = 3/6 We do the same for types B and C (multiply by 3) So : 6/9 * 5/8 * 4/7 * 3/6 * 3 = 5/14
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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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: At a blind taste competition a contestant is offered 3 cups of each of
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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 of each of
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15 Feb 2018, 05:47
To choose 2 types of sample out of 3, we get 3c2 We then have 6 cups and to choose 4 cups out of that we get 6c4. Hence, 3c2 * 6c4 The total ways of selecting 4 cups out of the total 9 is 9c4. Final ans is (3c2 * 6c4) / 9c4
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 Mar 2013, 02:32
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.
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 May 2017, 10:22
cledgard wrote: 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 1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9 Interesting way to approach the problem, though.



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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05 Jun 2019, 22:21
Although this approach may not be great, I used the elimination method after finding total possible ways of selecting 4 cups out of 9 = 4C9=112.
As I was not able to come up with the exact way to calculate the ways to select 4 cups of 2 samples...i checked for denominators being divisible by 112, as no answer choice can be correct if the denominator isn't a factor of 112. I was left with 1/2 and 5/14 as a potential answer choice, and by gut feeling the case looked to be of probability less than 50%(selecting 4 out of 9) hence excluded 1/2 and marked 5/14.
Though not best, such methods may come handy as a guessing strategy i feel.



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At a blind taste competition a contestant is offered 3 cups of each of
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05 Jul 2019, 10:15
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. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\) Really good problem. Let's break it down.  Does order matter? No. (i.e. Are we selecting or arranging? Problem says "random arrangement," this means selecting/combinations; all 3 cups of A are the same)  Is there replacement? No, unless specifically stated assume there is no replacement.  Is each grouping of them the same? Yes, for the denominator total (fundamental counting principle) but No for the numerator (we need to find 4 cups with a certain condition). There are 2 possibilities, either 2 types of tea or 3 types of tea, we can't get 1 type of tea because there's 4 selections and only 3 cups of each type.  The easiest way to approach is to find P(2 teas) = 1  P(3 teas). I made a chart using slot method: Ways to choose first, second, third, any tea type (3 * 2 * 1 * 1) * Number of cup choices (\(\frac{3}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\)) Step by step: 3 ways to choose 1st tea type * 3 cups out of 9 total 2 ways to choose 2nd tea type * 6 remaining cups out of 8 total 1 way to choose 3rd tea type * 3 remaining cups out of 7 total 1 way to choose ANY type * 6 remaining cups out of 6 total (anything can fulfil this requirement; we are asking for the way to choose ANY tea, rather than the way to choose one of the 3 types. Since we can choose any tea, we can choose any cup also) After reducing fractions, we are left with 9/14. Usually GMAT problems on probability will have answers that add up to 1, so I would expect that as a trap answer. Since we found the P(3 types) and are looking for P(2 types) we need to do 1  9/14 = 5/14. locklock wrote: I am having issues with the above, especially this line there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.
in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2? locklock That solution is similar to mine, but the poster reasoned through it a bit differently. I think what he meant was that there's 2 distinct ways for the second/third slot to account for (Say you pick Tea A for first slot, you can get B or C for 2nd slot. If you got B, then the 3rd is A, but if you got A, then the third is B). That's why he multiplied by 2. I would write it like this because it's more clear to me: Ways to choose ANY tea, 2nd tea, 3rd tea, ANY tea (1 * 2 * 1 * 1) * Number of choose cup choices ( \(\frac{9}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\)) Step by step: 1 way to choose ANY tea type * 9 cups out of 9 total 2 ways to choose 2nd tea type * 6 remaining cups out of 8 total 1 way to choose 3rd tea type * 3 remaining cups out of 7 total 1 way to choose ANY type * 6 remaining cups out of 6 total. You can see that the difference is only where the factor of 3 goes, whether in the ways to choose or the choices, so it works out to be the same. Does that make sense?



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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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: At a blind taste competition a contestant is offered 3 cups of each of
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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 of each of
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07 Apr 2016, 07:09
Is this an okay way to think about this problem?
In the case where we want to use the alternate solution (1opposite), would it be okay to say the following:
The number of ways to taste all the cups would result from the following choices:
(A A) B C
 Of the 3 samples, choose 1 to be a double  Of the 3 samples, choose 2 to be singles  Given our choice for the double, choose 1 location of 3 possible to represent the double  Given our choice for the singles, choose 2 locations of the 3 possible to represent the singles
Hence our total desired outcome would be: (3C1)(3C2)(3C1)(3C2) / (9C4) = 9/14. Our answer, therefore, would be 1  9/14 = 5/14



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At a blind taste competition a contestant is offered 3 cups of each of
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22 Feb 2019, 10:49
Can someone explain where I am going wrong?
This was my approach: For the first cup: 9 options For the second cup: 5 options For the third cup: 4 options For the fourth cup: 3 options
(9 * 5 * 4 * 3)/(9C4 * 3!)
I am using the 3! to remove the order




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