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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?

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

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 9-1 (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.

I think this is a poor-quality 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!

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.

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

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

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 : 1-9/14 = 5/14.

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

Would 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 \(\frac{6}{9}*\frac{5}{8}*\frac{4}{7}* \frac{1}{2}*3=\frac{5}{14}\)

I think this is a poor-quality 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.

I think this is a poor-quality 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.

Difficulty does no way means poor quality.
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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 (2-1-1).

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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-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- When nothing seem to help, I would go and look at a Stonecutter hammering away at his rock perhaps a hundred time without as much as a crack showing in it. Yet at the hundred and first blow it would split in two. And I knew it was not that blow that did it, But all that had gone Before.