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At a blind taste competition a contestant is offered 3 cups

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Manager
Joined: 29 Nov 2016
Posts: 83
At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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.

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.

$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.

Hope it's clear.
Intern
Joined: 17 Jul 2018
Posts: 19
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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.

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.

$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.

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?
Intern
Joined: 17 Jul 2018
Posts: 19
At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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.

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.

$$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$$.

Hope it's clear.

Intern
Joined: 11 Oct 2016
Posts: 13
GMAT 1: 680 Q48 V35
GPA: 3.61
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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23 Aug 2018, 07:51
Intern
Joined: 05 Mar 2015
Posts: 49
Location: Azerbaijan
GMAT 1: 530 Q42 V21
GMAT 2: 600 Q42 V31
GMAT 3: 700 Q47 V38
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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
Intern
Joined: 22 Dec 2016
Posts: 18
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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?
Intern
Status: Applying
Joined: 24 Oct 2017
Posts: 30
Location: India
WE: Investment Banking (Investment Banking)
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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04 Nov 2018, 01:24
Really interesting question. Are there any similiar questions, one could practice?
Intern
Joined: 07 Oct 2018
Posts: 6
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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?
Intern
Joined: 25 Aug 2018
Posts: 9
Location: India
Concentration: Strategy, Finance
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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18 Nov 2018, 03:06
What would be the difficulty level of this question on the exam?
VP
Joined: 07 Dec 2014
Posts: 1152
Re: At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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
Intern
Joined: 10 Aug 2018
Posts: 3
At a blind taste competition a contestant is offered 3 cups  [#permalink]

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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 = $$1 - \frac{9}{14} = \frac{5}{14}$$

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
At a blind taste competition a contestant is offered 3 cups &nbs [#permalink] 16 Dec 2018, 05:31

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