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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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29 Jan 2014, 12:22
Great explanations above. A rather headacheinducing 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





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At a blind taste competition a contestant is offered 3 cups [#permalink]
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28 Oct 2014, 23:39
# 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.
Hope this helps?



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At a blind taste competition a contestant is offered 3 cups [#permalink]
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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 [#permalink]
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24 Mar 2015, 02:45
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 Hence required P = 1 9/7 =
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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07 Sep 2015, 03:25
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!
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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19 Oct 2015, 00:04
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?
# \(\frac{1}{12}\) # \(\frac{5}{14}\) # \(\frac{4}{9}\) # \(\frac{1}{2}\) # \(\frac{2}{3}\) 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. Hi bunuel, Could you please explain this. C13  # of ways to choose the sample which will provide with 2 cups; ( here we are selecting 2 cups from 3 cups right?) Thanks.



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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19 Oct 2015, 08:52
guess it's probability= desired result/all the possible result?
cup1: it's 1, you'll taste a sample for sure as long as you take a cup
cup2: one sample is excluded and cup 1 is excluded, 4 cups left: 4/91
cup3, one sample is excluded and cup 1/2 are excluded, 3 cups left:3/81
cup 4, one sample is excluded and cup 1/2/3 are excluded, 2 cups left: 2/71
so the probability is
1*4/8*3/7*2/6=1/14



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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01 Nov 2015, 17:07
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!
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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23 Nov 2015, 19:06
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.



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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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 [#permalink]
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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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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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14 Apr 2017, 13:00
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. Bunuel please help me with this: My thought process: We have two possibilities : take two two cups from 2 samples or 3 cups from 1 sample and 1 cup from another sample 3C1 * 3C3 *2C1 *3C1 + 3C1*3C2*2C1*3C2 Divided by total no of cases 9c4 But I am getting answer 4/7 please tell where I am wrong



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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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 [#permalink]
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26 May 2017, 21:16
BunuelWhy is it that in your second method 1  (3C1*3C2*3C1*3C1)/9C4 , you do not have to divide (3C1*3C2*3C1*3C1) by 4! to avoid "duplication"/"doublecounting" since arrangement does not matter here? Thanks!



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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26 May 2017, 21:21
BunuelSorry, adding to my last post, I am confused because in the below question you posted, we need to divide the choices by the factorial of the number of groups: Q: How do divide 12 chocolates into 4 groups (groups are not distinct) A: (12C3*9C3*6C3*3C3)/4!



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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21 Jun 2017, 23:02
Can somone please guide me what is wrong with this approach.
For Denominator= 9C4
For numerator: 1. 3 cups from any one sample and 1 cup from any of the rest 2 samples = 3C1*3C3*2C1*3C1= 18 2. 2 cups from any one sample and 2 cups from any one sample of the rest 2 samples= 3C1*3C2*2C1*3C2=54



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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
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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 [#permalink]
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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 [#permalink]
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20 Feb 2018, 10:52
Why i am wrong with: 1 (3c1*3c1*3c1*6c1)/9c4 where 6c1 = 4rth cup can be chosen from any of the remaining (2,2,2 = 6 cups) Bunuel




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