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# At a blind taste competition a contestant is offered 3 cups

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

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06 Mar 2013, 01:32
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tosattam wrote:
Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth.
That gives a complete different perspective to the problem.
Where am I going wrong here?

A blind taste simply means that a contestant doesn't know which samples he/she is offered to taste.
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09 Mar 2013, 21:37
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Hi,
Let me try..
mainhoon, you are going with the logic that you select 1 cup from each sample in 3c1 x 3c1 x 3c1 ways and 1 remaining cup from 6 cups in 6c1 ways rite!
I guess the flaw with your approach is that this 6c1 is the probablity of selecting 1 item from 6 DIFFERENT items (r items from n different items is nCr)
Im this case the 6 items left are not different! They are XX,YY,ZZ types right! so you just cant use the 6c1 fomula!
This is the formula : The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of x^r in the expansion

hope it helps!

mainhoon wrote:
Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this?
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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29 Jan 2014, 10:59
If we employ the counting method applied in the question http://gmatclub.com/forum/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 ?
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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29 Jan 2014, 11:22
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
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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30 Jan 2014, 01:50
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?
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At a blind taste competition a contestant is offered 3 cups [#permalink]

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28 Oct 2014, 22:39
# 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.

Hope this helps?
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At a blind taste competition a contestant is offered 3 cups [#permalink]

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12 Jan 2015, 07:47
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, 01:45
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

Hence required P = 1- 9/7 =
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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07 Sep 2015, 02:25
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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18 Oct 2015, 23:04
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?

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

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.

Hi bunuel,

Could you please explain this.

C13 - # of ways to choose the sample which will provide with 2 cups; ( here we are selecting 2 cups from 3 cups right?)

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

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19 Oct 2015, 07:52
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guess it's probability= desired result/all the possible result?

cup1: it's 1, you'll taste a sample for sure as long as you take a cup

cup2: one sample is excluded and cup 1 is excluded, 4 cups left: 4/9-1

cup3, one sample is excluded and cup 1/2 are excluded, 3 cups left:3/8-1

cup 4, one sample is excluded and cup 1/2/3 are excluded, 2 cups left: 2/7-1

so the probability is

1*4/8*3/7*2/6=1/14
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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01 Nov 2015, 16:07
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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, 18:06
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.
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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01 Dec 2015, 21:45
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happyface101 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 = $$1 - \frac{9}{14} = \frac{5}{14}$$

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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17 Mar 2016, 23:58
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The probability of tasting all 3 cups is

One cup from pool A ($$\frac{3}{9}$$) x One cup from pool B ($$\frac{3}{8}$$) x One cup from pool C ($$\frac{3}{7}$$)

Since the contestant has to taste 4 cups, one possibility is picking second cup from pool A, so the probability is $$\frac{2}{6}$$

The probability of tasting in order of ABCA is $$\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6}$$ ..... (1)

In how many ways can the contestant taste ABCA cups = $$\frac{4!}{2!}$$ = 4 x 3 ....... (2)

The combinations ABCB & ABCC are also possible (along with ABCA) = 3 combinations ..... (3)

Since the choice of "a specific cup" from "a pool" doesn't matter here, we need NOT account for $$A_1A_2A_3, A_1A_3A_2,$$ ... etc. combinations

So Probability of tasting all three cups is = (1) * (2) * (3)

= $$\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6} * 4 *3 *3$$

=$$\frac{9}{14}$$

Hence, Probability of NOT tasting all three cups = $$1- \frac{9}{14} = \frac{5}{14}$$
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Re: At a blind taste competition a contestant is offered 3 cups [#permalink]

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07 Apr 2016, 06:09
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] 07 Apr 2016, 06:09

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