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At a blind taste competition a contestant is offered 3 cups

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At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 20 Feb 2018, 20:05
In the second approach why don't we multiply by \(C^1_2\) while selecting the second or third tea type?



Bunuel wrote:
Economist wrote:
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?

# \(\frac{1}{12}\)
# \(\frac{5}{14}\)
# \(\frac{4}{9}\)
# \(\frac{1}{2}\)
# \(\frac{2}{3}\)


And the good one again. +1 to Economist.

"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>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 cups each = 6 cups);
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

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 chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 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.

Hope it's clear.
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 12 Aug 2018, 18:07
Bunuel wrote:
Economist wrote:
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?

# \(\frac{1}{12}\)
# \(\frac{5}{14}\)
# \(\frac{4}{9}\)
# \(\frac{1}{2}\)
# \(\frac{2}{3}\)


And the good one again. +1 to Economist.

"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>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 cups each = 6 cups);
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

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 chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 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.

Hope it's clear.


\(C^1_3\) - # of ways to choose the sample which will provide with 2 cups;
\(C^2_3\) - # of ways to chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from third sample;
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

shouldn't this fraction be 12 / 126? How did we get 9 / 14?
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At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 12 Aug 2018, 18:09
Mudit27021988 wrote:
In the second approach why don't we multiply by \(C^1_2\) while selecting the second or third tea type?



Bunuel wrote:
Economist wrote:
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?

# \(\frac{1}{12}\)
# \(\frac{5}{14}\)
# \(\frac{4}{9}\)
# \(\frac{1}{2}\)
# \(\frac{2}{3}\)


And the good one again. +1 to Economist.



"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>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 cups each = 6 cups);
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

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 chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 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.

Hope it's clear.


Apologies, please ignore
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 23 Aug 2018, 07:51
ScottTargetTestPrep Please help on this one. Can you provide with an alternative solution?
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 25 Aug 2018, 23:43
I really dont understand the question. I dont try to solve, but just simply understand the situation. I can't
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 21 Oct 2018, 03:44
Hi Bunuel,
Sample: A B C
0 1 3- 6 ways
0 2 2-3 ways
1 2 1-3 ways
Total 12 ways to taste 4 samples out of 3*3 cup, 9 ways with 2 samples only, so probability shall be 9/12=3/4.
Where i am going wrong?
3 cups of every sample will be same. Isn't it?
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 04 Nov 2018, 01:24
Really interesting question. Are there any similiar questions, one could practice?
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 18 Nov 2018, 01:39
Hi everyone! I need some help

Here is how I started to solve this:

9/9*5/8*4/7*3/6=5/28

9/9 - the probability of choosing the first sample (we can choose any of those three)
5/8 - probability of choosing either 2 of the same samples as first or one cup of the second sample.
etc.

The logic is that starting from the second we keep three cups of the same sample away (we don't choose them)

What you think, is it right method and I just skipped a "2" somewhere or it is totally incorrect?
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 18 Nov 2018, 03:06
What would be the difficulty level of this question on the exam?
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Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 20 Nov 2018, 18:18
Economist wrote:
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. \(\frac{2}{3}\)


there are 9 possibilities for NOT tasting all 3 samples in 4 cups:
3A,1B
2A,2B
1A,3B
3B,1C
2B,2C
1B,3C
3C,1A
2C,2A
1C,3A
there are 3 possibilities for tasting all 3 samples in 4 cups:
1A,2B,1C
1A,2C,1B
2A,1B,1C
9/12=3/4 probability of not tasting all 3 samples
can someone please tell me what I'm missing here?
thank you,
gracie
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At a blind taste competition a contestant is offered 3 cups  [#permalink]

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New post 16 Dec 2018, 05:31
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\)
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\)
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\)
the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)



Please help EXPERTS!! :(

Could you please explain why it is multiplied by 2 (ways in which second & third cup can be tested)?
I solved this problem with an analogy to the couple problem (given below) and I missed this '2' thing and I didn't arrive to the correct answer.
My solution : 1-[(9/9)*(6/8)*(3/7)*(6/6)]=19/28


Couple problem: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
Solution: (12/12)∗(10/11)∗(8/10)∗(6/9)=48/99=16/33 <--- Ans
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