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Re D0138 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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Good day
I am unsure why this method will not work:
Choose any sample (9/9) x choose any other sample (8/8) x choose any of the samples of tea already chosen (4/7) and again as in previous choice, any of the samples already chosen (3/6) = 2/7 > Probability that I will choose only two different samples. Why does this not work?
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Last edited by ZAman1 on 12 Dec 2014, 05:24, edited 1 time in total.



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Re: D0138 [#permalink]
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12 Dec 2014, 05:20
I have worked on the problem a bit more and now just have more questions!
Prob of selecting 3 of 1 one sample: (3/9*2/8*1/7)*3/6 = 1/168 but can select either samples A,B or A,C or B,C and can also select in different orders, i.e. AAAB; AABA; ABAA; BAAA therefore prob of selecting 3 of one sample = 1/168 x 2C3 x (4!/3!1!) = 1/14
If we follow the same principle for selecting 2 of each samples: (3/9*2/8*3/7*2/6) = 1/84 but can select either sample A,B or A,C or B,C and can also select in different orders, i.e. AABB; ABAB; ABBA; BAAB; BABA; BBAA therefore prob of selecting 2 of each sample = 1/84 x 2C3 x (4!/2!2!) = 3/14
Total probability equals 4/14 (1/14 + 3/14). What am I missing!!!!



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My apologies for bumping my own thread, but just want to show how prob could work. I still have no idea why the other methods does not work: Prob of selecting sample A&B: (6/9*5/8*4/7*3/6)=5/42 but could also select A&C or B&C > 5/42 x 2C3 = 5/14 WHY!!!



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Re: D0138 [#permalink]
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12 Dec 2014, 06:54
ZAman1 wrote: I have worked on the problem a bit more and now just have more questions!
Prob of selecting 3 of 1 one sample: (3/9*2/8*1/7)*3/6 = 1/168 but can select either samples A,B or A,C or B,C and can also select in different orders, i.e. AAAB; AABA; ABAA; BAAA therefore prob of selecting 3 of one sample = 1/168 x 2C3 x (4!/3!1!) = 1/14
If we follow the same principle for selecting 2 of each samples: (3/9*2/8*3/7*2/6) = 1/84 but can select either sample A,B or A,C or B,C and can also select in different orders, i.e. AABB; ABAB; ABBA; BAAB; BABA; BBAA therefore prob of selecting 2 of each sample = 1/84 x 2C3 x (4!/2!2!) = 3/14
Total probability equals 4/14 (1/14 + 3/14). What am I missing!!!! There is a long discussion on this question 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 [#permalink]
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12 Feb 2015, 09:48
I did it like this: If 3 types of cups (A,B,C), Cups to be selected: ABBB AABB AAAB Probability will be: (3/9*3/8*2/7*1/6) for ABBB = 1/168 (3/9*2/8*3/7*2/6) for AABB = 1/84 (3/9*2/8*1/7*3/6) for AAAB = 1/168 Adding all: 1/42! Sorry if I did it wrong. I am not so good in probability!
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Re D0138 [#permalink]
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10 Aug 2015, 20:12
I think this is a highquality question and the explanation isn't clear enough, please elaborate. I agree with the fact that you chose the sample with 2 cups in 3c1 ways but then why are we not choosing the sample with 1 cup in a similar manner. Why not do this? 3c1*3c2*(for second sample2c1*3c1)*(third sample1c1*3c1)?? Please explain!!



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Re: D0138 [#permalink]
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17 Aug 2015, 03:35



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Re D0138 [#permalink]
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07 Feb 2016, 10:05
I think this is a poorquality question and the explanation isn't clear enough, please elaborate.



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Re: D0138 [#permalink]
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07 Feb 2016, 10:06



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Re: D0138 [#permalink]
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15 Feb 2016, 09:59
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 Sorry Bunuel but didnt get this part. "C46C64  # of ways to choose 4 cups out of 6 cups of two samples ( 2samples∗3cups each=6cups2samples∗3cups each=6cups )" Can you please explain the concept behind this application. Got the rest correct. Am a beginner in Probability questions.



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Re D0138 [#permalink]
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08 Mar 2016, 18:23
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. please show how you worked out the solution, how did you get the 5/14. I looked at combinatorics lessons online but cannot find this result. Thank you



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Re D0138 [#permalink]
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09 Mar 2016, 05:50
I think this is a highquality question.



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Re: D0138 [#permalink]
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09 Mar 2016, 05:55
Soukaina92 wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. please show how you worked out the solution, how did you get the 5/14. I looked at combinatorics lessons online but cannot find this result. Thank you Please look at the solution at d01183499.html#p1413648 before asking the question. Where exactly are finding a problem with the solution? It is clear and concise.



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Re: D0138 [#permalink]
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09 Mar 2016, 06:45
I have looked at the solution already. The result I am getting is 3/10 so it must be the way I am working out the factorial. Is it possible to post a solution step by step, with all the calculations? Thanks
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Re: D0138 [#permalink]
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09 Mar 2016, 07:24
Soukaina92 wrote: I have looked at the solution already. The result I am getting is 3/10 so it must be the way I am working out the factorial. Is it possible to post a solution step by step, with all the calculations? Thanks
Posted from my mobile device Can you show your steps that you have employed to arrive at the solution of 3/10, this way we will be able to see where exactly are you making a mistake.







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