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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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Re D0138
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16 Sep 2014, 00:13
Official Solution: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. \(\frac{2}{3}\) "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 \(\gt\)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 (\(2 samples*3 \text{cups each} = 6 cups\)); \(C^4_9\)  total # of ways to choose 4 cups out of 9. 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 choose these 2 cups from the chosen sample; \(C^1_3\)  # of ways to choose 1 cup out of 3 from second sample; \(C^1_3\)  # of ways to choose 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
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Re: D0138
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20 Nov 2014, 21:45
Bunuel wrote: Official Solution: \(C^4_9\)  total # of ways to choose 4 cups out of 9.
Hi Bunuel, I'm having trouble understanding the format you are using in your explanation. In the combinatorics thread in your signature, it says: A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: \(C^n_k\) So in your explanation, this would mean choosing 9 objects out of only 4 distinct objects. How is this possible? We would be getting a fraction here when we should be getting whole numbers.



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Re: D0138
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21 Nov 2014, 04:52
OutOfTheHills wrote: Bunuel wrote: Official Solution: \(C^4_9\)  total # of ways to choose 4 cups out of 9.
Hi Bunuel, I'm having trouble understanding the format you are using in your explanation. In the combinatorics thread in your signature, it says: A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: \(C^n_k\) So in your explanation, this would mean choosing 9 objects out of only 4 distinct objects. How is this possible? We would be getting a fraction here when we should be getting whole numbers. \(C^4_9\), \(C^9_4\), 9C4 are the same: choosing 4 out of 9. How can we choose 9 out of 4? Those are just different ways of writing the same.
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For other solutions check long and good discussion HERE. Hope it helps.
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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: D0138
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30 May 2016, 14:43
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?
So, we have : Sample of tea A : the cups are A1, A2, and A3 Sample of tea B : the cups are B1, B2, and B3 Sample of tea C : the cups are C1, C2, and C3
So the contestant is offered : 4 random cups from the set {A1, A2, A3, B1, B2, B3, C1, C2, C3}
We want to calculate the probability to not choose the 3 samples with 4 cups ; for instance A1A2A3B2, or B13BC1C2, or C2A1A2A3, or A1A2B2B3, etc. >So the contestant could test ONLY 2 of the 3 samples
Proba = \(\frac{(Number.of.ways.to.choose.4.cups.from.2.samples)}{(Number.total.of.ways.to.choose.4.cups.from.3.samples)}\)
1/ Number of ways to choose 4 cups from 2 samples :
Stage 1 : Choose 2 samples from the 3 = \(\frac{3*2}{2}\) = 3 We obtain : A, A, A, B, B, B for example
Stage 2 : Choose 4 from the 6 cups selected = \(\frac{6*5*4*3}{4*3*2}\) = 15
Counting Principle : 3*15 ways
2/ Number of ways to choose 4 cups from 3 samples = from 9 cups :
\(\frac{9*8*7*6}{4*3*2}\) = 9*7*2
So, proba = \(\frac{3*15}{9*7*2}\) = \(\frac{5}{14}\)



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Re: D0138
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30 Jun 2016, 08:23
My "layman" approach: 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?Simply calculating \(\frac{#possible.cases}{#tot. cases}\) we can find the solution quite fast. First possible cases: \(9*5*4*3\) We have 9 different ways in which we can taste the first cup of tea, after this first one we we do 91 (last draw) 3 one family of teas, so the following draws are 5, 4 and 3. Total cases: \(9*8*7*6\) Nothing surprising here. \(\frac{(9*5*4*3)}{(9*8*7*6)}\) = \(\frac{5}{28}\) we still have to multiply * 2 because there are 2 ways in which the 2nd the 3rd cup can be drawn > AB and BA. So \(\frac{5}{28} * 2\) = \(\frac{5}{14}\)



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Re D0138
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12 Jul 2016, 12:01
I think this is a highquality question.



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IMO this is a great question.
Here is my approach:
Given: 3 Samples of tea and for each sample we have 3 cups.
Samples of tea : Cups of tea A : cup1, cup2, cup3 B : cup4, cup5, cup6 C : cup7, cup8, cup9
Contestant tastes 4 different cups of tea ==> It means he will taste any 4 cups from the 9 cups.
We are asked to find the probability that "a contestant does not taste all of the samples".
"a contestant does not taste all of the samples"> means contestant have to taste any 2 samples out of the 3 (A,B,C).
Contestant can taste samples as i) A or B ii) B or C iii) A or C
How many ways contestant can does so => Select 4 cups out of 6 cups (\(6C4\)) ==> 15 ways It can be done in 3 ways (AB, BC, or AC)
==> Total number of ways a contestant does not taste all of the samples = 3 * 15 = 45
Total ways to select 4 cups out of 9 cups = \(9C4\) = 126
Proability = 45/126 = 5/14



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Re: D0138
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25 Feb 2017, 12:00
Here is a simple solution: P(does not taste all samples) = 1  P(does taste all samples) P(does taste all samples) means he tastes A, B, C, and 1 more (e.g. ABCA, ABCB, ABCC) Find the probability of one happening P(ABCA) = 3/9 * 3/8 * 3/7 * 2/6 = 1/56 Then multiply by the # of ways to arrange each scenario: 4!/2! = 12 (mississippi rule) Then multiply by 3 to account for all three scenarios (1/56)*(12)*(3) = 9/14 Then subtract from 1 to get the answer: 1(9/14) = 5/14
Answer B



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Re: D0138
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06 Apr 2017, 10:56
lets consider the case of the contestant tasting all the samples ; Cases : AABC ,ABBC , ABCC For each case the selection is 3c2x3c1x3c1 : so, total is 3X3c2x3c1x3c1 Now the probability of contestant to taste all the three samples is (3X3c2x3c1x3c1)/(9c4) = 9/14 the probability that a contestant does not taste all of the samples : 19/14 = 5/14.



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Re: D0138
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17 Apr 2017, 11:52
This alternative solution of filling spaces method also helps:
We have 9 cups in total, 3 cups per each of the 3 samples.
It is an selection question (order does not matter, irrelevant way we choose the samples)
Number of ways of choosing 4 cups of all the samples: (9 x 8 x 6 x 3) / (1 x 2 x 3 x 4) =54=9 x 6
Number of ways of choosing 4 cups (9 x 8 x 7 x 6) / (1 x 2 x 3 x 4) =9 x 14
Probability of not tasting all samples = 1  probability tasting all samples
= 1 (9x6 / 9x14) = 5/14



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Re: D0138
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04 Jun 2017, 07:58
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 cledgardWould not be the probability of not choosing 1st cup Type A = 6/9? (where 6=3 of B + 3 of C, and 9= total # of cups) 2nd, not Type A = 5/8 3rd, not Type A = 4/7 4th, not Type A = 1/2 since 3/6 and 3 options to start with A, B, or C so we would get \(\frac{6}{9}*\frac{5}{8}*\frac{4}{7}* \frac{1}{2}*3=\frac{5}{14}\)



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1probability of tasting all types = probability of not tasting all types ways to have tasted all types = AABC, BBAC, CCAB i.e 3. also note we can rearrange each in 4!/2! ways. therefore total number of options = 3x4x3 now calculate the probability of AABC = 3/9 x 2/8 x 3/7 x 3/6 note from above that this particular event i.e. AABC can occur 3x4x3 ways therefore multiply this with the probability calculated for AABC = 9/14 therefore, subtract 9/14 from 1 = 5/14



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Re D0138
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20 May 2018, 08:43
I think this is a highquality question and I agree with explanation.
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I think here is the easiest approach for anyone to nail this question.
We have 3 groups of teas, let's call them Group A, B, C. As we need to find the probability of tasting only 2 groups in 4 picks, then in each picking streak we are ignoring one group.
So here is the multiplication:
First picking streak First pick: 6/9 (because 1 group is not participating  let's say we overlook group C, we have 6 choices out of 9) Second pick: 5/8 (we keep ignoring group C, and we have 5 choices left as 1 has been picked) 3rd pick: 4/7 4th pick: 3/6 6/9 x 5/8 x 4/7 x 3/6 = Probability of picking just 2 groups in 4 picks ignoring group C. We need to add another picking streaks as we have 2 groups left so you just multiple all that probability by 3.
6/9 x 5/8 x 4/7 x 3/6 x 3 = 5/14
If you think this is helpful, please give KUDOS. Thank you.



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BunuelRegarding the other way to solve this problem: Calculating the opposite probability and then substracting the same from 1, Here is my logic: First we select one cup from each of the r different varieties: 3C1 X 3C1 X 3C1 For the fourth cup we can choose one from any of the remaining 6 cups : 6C1 ways. Total no of events is 9C4 So the reqd probability for the opp event stands at (3C1 X 3C1 X 3C1)/9C4.... But this is >1... Please explain where am i going wrong Thanks



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Re: D0138
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05 Apr 2019, 23:26
One simple solution from my side: Suppose the cups are like A,A,A,B,B,B,C,C,C => 3A, 3B, 3C Not selecting all the 3 types of samples means selecting only 2 types of samples.
Now, possible ways of doing so is either 2A + 2B or 3A + 1B OR 1A + 3B.
multuply this by 3 as we can select A &B or A & C or B & C.
Solving this, we get [u]3C1 X 3C2 + 3C3 X 3C1 + 3C1 X 3C3/9C4 = 5/14



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Re D0138
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21 Apr 2019, 10:50
I think this is a highquality question and I agree with explanation.
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