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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. \(\frac{2}{3}\)

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 (2-1-1).

\(C^1_3\) - # 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.

\(C^4_9\) - 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.

\(C^4_9\) - 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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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?

Regards

Last edited by ZAman1 on 12 Dec 2014, 05:24, edited 1 time in total.

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!!!!

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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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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KUDOS please!! If it helped. Warm Regards. Visit My Blog

I think this is a high-quality 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 sample-2c1*3c1)*(third sample-1c1*3c1)?? Please explain!!

I think this is a high-quality 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 sample-2c1*3c1)*(third sample-1c1*3c1)?? Please explain!!

Hope alternative solution HERE help to understand better.
_________________

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 (2-1-1).

\(C^1_3\) - # 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.

"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.

I think this is a poor-quality 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

I think this is a poor-quality 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 d01-183499.html#p1413648 before asking the question. Where exactly are finding a problem with the solution? It is clear and concise.
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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

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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