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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]  06 Mar 2013, 01:32
Expert's post
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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Re: Diagnostic Question [#permalink]  09 Mar 2013, 21:37
1
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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]  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]  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]  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]  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]  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]  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] 24 Mar 2015, 01:45

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