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At a blind taste competition a contestant is offered 3 cups of each of

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

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New post 05 Jul 2019, 10:15
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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. \(\frac{2}{3}\)


Really good problem. Let's break it down.

-- Does order matter? No. (i.e. Are we selecting or arranging? Problem says "random arrangement," this means selecting/combinations; all 3 cups of A are the same)
-- Is there replacement? No, unless specifically stated assume there is no replacement.
-- Is each grouping of them the same? Yes, for the denominator total (fundamental counting principle) but No for the numerator (we need to find 4 cups with a certain condition). There are 2 possibilities, either 2 types of tea or 3 types of tea, we can't get 1 type of tea because there's 4 selections and only 3 cups of each type.
-- The easiest way to approach is to find P(2 teas) = 1 - P(3 teas). I made a chart using slot method:

Ways to choose first, second, third, any tea type (3 * 2 * 1 * 1) * Number of cup choices (\(\frac{3}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))

Step by step:
3 ways to choose 1st tea type * 3 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total (anything can fulfil this requirement; we are asking for the way to choose ANY tea, rather than the way to choose one of the 3 types. Since we can choose any tea, we can choose any cup also)

After reducing fractions, we are left with 9/14. Usually GMAT problems on probability will have answers that add up to 1, so I would expect that as a trap answer. Since we found the P(3 types) and are looking for P(2 types) we need to do 1 - 9/14 = 5/14.


locklock wrote:
I am having issues with the above, especially this line
there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?


locklock That solution is similar to mine, but the poster reasoned through it a bit differently. I think what he meant was that there's 2 distinct ways for the second/third slot to account for (Say you pick Tea A for first slot, you can get B or C for 2nd slot. If you got B, then the 3rd is A, but if you got A, then the third is B). That's why he multiplied by 2.

I would write it like this because it's more clear to me:

Ways to choose ANY tea, 2nd tea, 3rd tea, ANY tea (1 * 2 * 1 * 1) * Number of choose cup choices ( \(\frac{9}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))

Step by step:
1 way to choose ANY tea type * 9 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total.

You can see that the difference is only where the factor of 3 goes, whether in the ways to choose or the choices, so it works out to be the same.

Does that make sense?
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Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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New post 28 Jul 2019, 12:09
AKProdigy87 wrote:
I got B: 5/14 as well, although using a different approach.

Number of ways to choose 4 cups to drink from: 9C4

Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.

Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2

Three cases:

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3

Therefore,

\(P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}\)

\(P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}\)

\(P = \frac{45}126\)

\(P = \frac{5}{14}\)


this is how i tried to do in a way, however, want to know between a and c, if i multiple by 3 to consider any flavor, why do i need another combination of it.

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1 * 3 for any flavor
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2 *3 for any flavor
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3 -> Why do i need combination C
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Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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New post 16 Sep 2019, 08:38
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. \(\frac{2}{3}\)




Breakdown: There are 3 cups with three different samples and there are total of 9 cups .
So all 9 cups we can say ,
Sample 1--contained in 3 similar cups.
Sample 2--contained in 3 similar cups
Sample 2--contained in 3 similar cups

Total number of ways in which one can taste all the three samples(from a set of 4)
= 3C1(sample 1,one cup) * 3C1(sample 2,1cup)* 3C2(sample 3,2 cup) *3(total ways in which these three selections can be made(3!/2)=3)
=3*3*3*3= 81 ways

Required probability = 1 - 81/9c4 = 5/14
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Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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New post 03 Nov 2019, 07:47
locklock wrote:
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 = \(1 - \frac{9}{14} = \frac{5}{14}\)



I am having issues with the above, especially this line
there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?




please explain why we multiplied by 2 I cannot relate it with other examples such as https://gmatclub.com/forum/as-part-of-a ... 56446.html
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At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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New post 07 Nov 2019, 21:27
locklock wrote:
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 = \(1 - \frac{9}{14} = \frac{5}{14}\)



I am having issues with the above, especially this line
there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?



Bunuel, Gladiator59, generis - I have been racking my head over why that 2 is added. Please help us gain clarity.
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At a blind taste competition a contestant is offered 3 cups of each of   [#permalink] 07 Nov 2019, 21:27

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