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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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Hi,
Can somebody explain me why in this question we havent taken as such ((3C2*6C2) + (3C1*6C3) +(6C4))/9C4



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Re D0138
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09 Sep 2016, 21:40
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Im sorry... but I cant quite understand the formula you´re using bunuel =S could someone please explain me step by step with the solutions please? Thank you!



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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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27 Oct 2016, 08:50
Bunuel wrote: 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 Hi Bunuel  I still having a doubt In the opposite event, since you have calculated C(1,3) for the first sample where you are picking two cups  dont you need to calculate then C(1,2) for the second sample (you are selecting one group within two) and C(1,1) for the third? I know the third calculation is pointless (C(1,1)) but just for continue with the notation Thanks V



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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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Re D0138
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27 Jun 2017, 23:10
I think this is a poorquality question. " a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. " CONVERTING THIS INTO A MATHEMATICAL SITU WAS DIFFICULT.



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Re: D0138
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27 Jun 2017, 23:11



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Re: D0138
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30 Jul 2017, 23:05
Hello, Can anyone please let me know where am I doing wrong. 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). So prob = [9c4 3c1*3c1*3c1*6c1] / 9c4 (6c1 as I can choose any one out of the remaining 6 cups)
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Re: D0138
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23 Aug 2017, 13:50
Can some one explain what is wrong with my approach ??
Choose 1 cup of each sample = 3C1, 3C1, 3C1
Now, 6 cups are left so choose any 1 from the left 6 cups = 6C1
total no of probability is 9C4
(3*3*3*6) / 9C4
Can some one explain where I have made the mistake ?



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Re D0138
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28 Oct 2017, 21:46
I think this is a highquality question and I agree with explanation. Hi Bunnuel, In your Alternative2: We select one cups from each of the three samples first in 3c1*3c1*3c1 ways. Now 6 cups are remaining, and we have to select any one from it. It can be done in 6c1 ways. so the numerator becomes: 3c1*3c1*3c1*6c1 as opposed to your solution 3c1*3c1*3c2*3c1 Where am I wrong here??



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Re: D0138
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25 Nov 2017, 06:38
Bunuel wrote: Official Solution: 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 Bunuel, Does this approach make sense? \(C^1_3\)  # 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; \(C^2_3\)  # of ways to choosing 2 samples that will provide one cup each; \(C^1_3\)  # of ways to choose 1 cup out of 3 from the each of the previous samples; \(C^4_9\)  total # of ways to choose 4 cups out of 9. The result that I get is the same, but I know it may be because of this specific example, I want to know if my logic is right or not



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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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13 Mar 2018, 08:09
I don't really understand the answer to this question as I am out by a factor of 3 on the number of allowed combinations and can't find any more.
Total ways, NOT discounting duplicates, is 9!/5! (ie. 9 x 8 x 7 x 6). However, order doesn't matter (ie ABCA = ACBA) so we need to remove duplicates . Therefore we must divide again by 4!, giving us the required 9!/4!. Total ways is 9!/5!4! = 9x8x7x6 / 4x3x2x1.... cancels down to 9*7*2
Ways we can not try all 3 ie. ways we can try 2 only  AAA with any out of B1, B2, C1, C2 = 4 options BBB with any out of A1, A2, C1, C2 = 4 options CCC with any out of A1, A2, B1, B2 = 4 options total 12 options PLUS AABB, AACC, BBCC total 15 options. We don't need to consider the orders of the cups as they don't matter
However, to get to the requisite 5/14, we are out on the numerator by a factor of 3. I currently have 3*5/9*7*2 which simplifies to 5/42. Can anyone correct? How to get enough "allowed" combinations to cancel everything down to 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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