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

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29 Jan 2014, 12:22

Great explanations above. A rather headache-inducing question, I admit I had to resolve to the following guessing pattern after I passed the 1:36 mark and admitted defeat.

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. \(\frac{1}{12}\) Too low B. \(\frac{5}{14}\) Looks reasonable C. \(\frac{4}{9}\) Impossible, cannot be that simple D. \(\frac{1}{2}\) Too high E. \(\frac{2}{3}\) Too high

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

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30 Jan 2014, 02:50

Quote:

If we employ the counting method applied in the question tough-p-n-c-92675.html then Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another i.e. AAAB The number of ways this can happen is 3C2*4!/3! = 12 3C2 = Ways to select any 2 samples out of the 3 choices 4!/3! = # of ways AAAB can be rearranged

or 2. Contestant tastes 2 cups of a sample and 2 of another i.e. AABB The number of ways this could happen is 3C2*4!/(2!*2!) = 18 3C2 = Ways to select any 2 samples out of the 3 choices 4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126 = 5/21

Is this approach not correct? Where am I missing ?

I have come up with another approach which gives a different result again.

Assuming the that the contestant does not taste the 3rd sample there are 2 scenarios possible

AABB and AAAB

Counting one at a time. #of ways AABB can occur = 3C2*3C2*3C2 = 27 here, 3C2 = ways to select the 2 samples from 3 3C2 = ways to select the 2 cups from 3 cups of first sample 3C2 = ways to select 2 cups from 3 cups of second sample

# of ways AAAB can occur = 3C2*3C3*3C1 = 9 3C2 = ways to select the 2 samples from 3 3C3 = ways to select 3 cups from the 3 cups of first sample 3C1 = ways to select 1 cup from the 3 cups of second sample

Hence the total number of ways in which the contestant cannot taste any cup from third sample = 9 +27 = 36

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 36/126 = 4/14

Even after hours and hours of going through various material on Probablity and combinations I am struggling to identify a uniform approach that can be applied across problems. How would one know which approach to apply for which problem during the exams. Is there any way we can determine that?

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

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28 Oct 2014, 23:39

# of ways of choosing 4 cups from 9 = 9c4 = 126 # of ways of choosing 4 cups such that one gets only 2 varieties of tea -

1st, 2nd and 3rd cup to be from same type = 3*2*1 per type * 3 (for 3 different types) = 27 4th cup to be from a different type (can't be from same type as none of 1 type are left) = 3*3 per type * 2 (for 2 different types) = 18

Total = 1st, 2nd, 3rd & 4th cup = 27+18 = 45

Prob (Tastes only 2 varieties) = 45/126 = 5/14 - B.

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

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12 Jan 2015, 08:47

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

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24 Mar 2015, 02:45

Hi Is the following approach not correct..?? where am wrong????

C^1_3 - # of ways to chose 1 cup out of 3 from 1st sample; C^1_3 - # of ways to chose 1 cup out of 3 from 2nd sample; C^1_3 - # of ways to chose 1 cup out of 3 from third sample; C^1_6- # of ways to chose 1 cup out of 6 remaining cups;

C^4_9- # of ways to chose 4 cup out of 9 cups;

Probability = [C^1_3 x C^1_3 x C^1_3 x C^1_6] / C^4_9 = 162/126 =9/7

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

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07 Sep 2015, 03:25

Hi,

Can someone help me understand the question? I am having great trouble figuring out what exactly is happening here, let alone what's asked.

"3 cups of each of the 3 samples of tea" - does this mean that there are 3 types of tea, say Aasam Tea, Green Tea and Ginger Tea and a contestant is given 1 cup of each?

"in a random arrangement of 9 marked cups" - what does this mean? Are there 9 cups, each with a different "type" of tea, or are there just 9 cups that may/may not have tea? The three cups offered above - these come from these nine cups?

"each contestant tastes 4 different cups of tea" - each contestant as against "a" contestant? So what's happening here? Each tastes 4 cups, 3 of which he has been given as part of the first point above (all three of a different flavour + one he picks up on his own?)

"contestant does not taste all of the samples" - so probability that he does not taste each of aasam,green and ginger? As in, a contestant tastes either zero or one or two? What's the difference here between a contestant and each contestant?

I am completely confused! Can someone help explain what's happening with a diagram? Thanks in advance!
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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19 Oct 2015, 00:04

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?

"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.

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

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01 Nov 2015, 17:07

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

Hi! Can some please explain the note regarding "there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1."? I didn't understand why we need to multiple by 2.

Separately, why is my approach below wrong?

P= 1-(9*6*3*6)/(9 choose 4) is obviously wrong, but why? The denominator I understand very clearly - number of ways to choose 4 from 9 choices. The numerator is based on the reasoning that the first cup can be from any of the 9 choices, so 9 ways. The second cup has to be from any of the 6 cups (the two samples different from the first cup). The third cup has to be from any of the 3 cups (the third sample not choose by cup 1 or 2). And the last cup can be from any of the remaining 6 cups.

Thanks so much!
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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23 Nov 2015, 19:06

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

Hi Bunnuel, Is this a correct solution? I am having a difficult time understanding why did he multiply by two. I would expect that through this method, you don't need to multiply to account for combinations since the probabilities for each pick already account for that. Thanks.

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

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07 Apr 2016, 07:09

Is this an okay way to think about this problem?

In the case where we want to use the alternate solution (1-opposite), would it be okay to say the following:

The number of ways to taste all the cups would result from the following choices:

(A A) B C

- Of the 3 samples, choose 1 to be a double - Of the 3 samples, choose 2 to be singles - Given our choice for the double, choose 1 location of 3 possible to represent the double - Given our choice for the singles, choose 2 locations of the 3 possible to represent the singles

Hence our total desired outcome would be: (3C1)(3C2)(3C1)(3C2) / (9C4) = 9/14. Our answer, therefore, would be 1 - 9/14 = 5/14

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

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

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14 Apr 2017, 13:00

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?

"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.

Bunuel please help me with this: My thought process: We have two possibilities : take two two cups from 2 samples or 3 cups from 1 sample and 1 cup from another sample

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

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06 May 2017, 10:22

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cledgard wrote:

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

1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9

Why is it that in your second method 1 - (3C1*3C2*3C1*3C1)/9C4 , you do not have to divide (3C1*3C2*3C1*3C1) by 4! to avoid "duplication"/"double-counting" since arrangement does not matter here? Thanks!

Sorry, adding to my last post, I am confused because in the below question you posted, we need to divide the choices by the factorial of the number of groups:

Q: How do divide 12 chocolates into 4 groups (groups are not distinct) A: (12C3*9C3*6C3*3C3)/4!

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

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21 Jun 2017, 23:02

Can somone please guide me what is wrong with this approach.

For Denominator= 9C4

For numerator:- 1. 3 cups from any one sample and 1 cup from any of the rest 2 samples = 3C1*3C3*2C1*3C1= 18 2. 2 cups from any one sample and 2 cups from any one sample of the rest 2 samples= 3C1*3C2*2C1*3C2=54

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

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25 Sep 2017, 11:15

DanceWithFire wrote:

cledgard wrote:

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

1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9

Interesting way to approach the problem, though.

You are right, I made a mistake when I wrote it no the forum. The correct approach is the following: I get the probability that each cup is not a specific type of tea (type A): 1st cup, Probability it is not type A = 6/9 2nd cup, Probability it is not type A = 5/8, since we only have 8 cups left, and 3 of them are type A 3rd cup, Probability it is not type A =4/7 4th cup, Probability it is not type A = 3/6 We do the same for types B and C (multiply by 3) So : 6/9 * 5/8 * 4/7 * 3/6 * 3 = 5/14
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