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At a blind taste competition a contestant is offered 3 cups [#permalink]
14 Nov 2009, 06:34
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00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
59% (03:05) correct
41% (02:10) wrong based on 226 sessions
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. \(\frac{2}{3}\)
Re: Prob- gclub diagonostics [#permalink]
14 Nov 2009, 07:06
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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?
"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: Prob- gclub diagonostics [#permalink]
15 Nov 2009, 03:16
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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
Diagnostic Question [#permalink]
15 Aug 2010, 16:17
I thought the answer to this question can be obtained thus: 1 - P(all samples are tasted)
P(all samples tasted) = I choose one from each of the 3 samples = 3C1 x 3C1x 3C1 x 6C1 as there will be 6 cups left from which I can get the 4th cup = 27 x6 = 162 makes no sense as there are only 9C4 combinations = 126.
S1 A B C S2 D E F S3 G H I
Select 1 from S1 in 3 ways, same with S2, same with S3 = a total of 27 ways I have 6 cups left, can select 1 in 6 ways
So total of 27 x 6 = 162 ways. It seems that this should be 81, I am double counting somewhere. Only 81 gives me the right answer...
Re: Diagnostic Question [#permalink]
15 Aug 2010, 17:43
It should be 1-(3*3C2*3C1*3C1/9C4) = 1-9/14 = 5/14
Now why is your approach wrong ? E.g: in how many ways 2 objects can be selected from a set of 3. We know it is 3C2 As per the approach you took above it should be equal to; 3C1*2C1 != 3C2.
Re: Diagnostic Question [#permalink]
15 Aug 2010, 18:03
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We want to select 2 cups of tea which are of similar type and the rest 2 cups which each are of different type ; hence the terms 3C2 * 3C1* 3C1 Now you should multiply it by 3 because ; out of the 3 tea types , you may select the tea type which is in 2 cups, in 3 ways.
Re: Diagnostic Question [#permalink]
15 Aug 2010, 18:16
Maybe I dont understand the question. But I was approaching it as 1 - p(he tastes all samples), so why limit to just 2 cups and then make it so that the other 2 are each of a different type?
S1 A B C S2 D E F S3 G H I
So why can't be go ADGF - select one from S1, one from S2, one from S3 and then have 1 of the remaining 6? You are doing 3c2 3c1 3c1, so that is picking 2 from one row and one each from the other 2 rows such as ABDG. Perhaps if you can explain how to get an accurate count of combinations, when I select 1 from each row, make it 3 cups and then pick the 4th one of the remaining 6? _________________
Re: Diagnostic Question [#permalink]
15 Aug 2010, 18:53
I am not sure if it could be calculated that way. Lets make it even simpler S1 A B C
In how many ways can you select 2 out of these 3 ? 3C2, right ? If however; if I first select 1 of them say in 3C1 ways; there are 2 still left. Now, if I again select 1 out of the remaining two, in 2C1 ways, the total no. of combination 3C1*2C1, which is not equal to 3C2.
Re: Diagnostic Question [#permalink]
15 Aug 2010, 19:06
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? _________________
Re: Prob- gclub diagonostics [#permalink]
17 Aug 2010, 00:53
1
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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
I don't follow this answer. (a) and (c) show that he ends up tasting all the cups of the given sample, we are not supposed to allow this to happen? _________________
Re: Prob- gclub diagonostics [#permalink]
10 Nov 2010, 00:11
I was all over the place in trying to solve this...neways Thanks Bunuel..that was a magnificient and clear solution..but my only concern is on the actual GMAT..is this question easy enuf to be answered in under 2 mins...???or rather...it looks like a difficulty level 800+ to me...
Re: Prob- gclub diagonostics [#permalink]
06 Feb 2011, 07:18
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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 = \(1 - \frac{9}{14} = \frac{5}{14}\)
_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Re: Prob- gclub diagonostics [#permalink]
03 Aug 2011, 23:47
Can anyone tell me where I went wrong? I understand you can derive the answer from the solution above, but why does my method below give me a wrong answer?
I used 1 - (prob of drinking all three teas) Total combinations = 9C4 = 126 prob of drinking all threes = 1 of each tea + 1 of any tea. Using A, B, & C to represent the teas: Combinations where all three teas are represented (2 of one tea and one of each of the other tea) = A, A, B, C A, B, B, C A, B, C, C
Combinations of rearranging A,A,B,C = 4!/2! = 12 Combinations of rearranging A,B,B,C = 4!/2! = 12 Combinations of rearranging A,B,C,C = 4!/2! = 12
Total combinations of drinking each tea = 36 Probability of not drinking all tea = 1 - (36/126) = 90/126 = 5/7
I can't figure out where the logical error or where I double counted or ignored possibilities come from. Can anyone help?
Re: Prob- gclub diagonostics [#permalink]
22 Dec 2011, 22:41
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.
Thank you. If the question was about the probability that a contestant does not taste all the 3 samples of a cup. (That's what I wrongly understood from this question first). Then the answer would be \(\frac{6}{7}\)?
Re: At a blind taste competition a contestant is offered 3 cups [#permalink]
05 Mar 2013, 08:23
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?
gmatclubot
Re: At a blind taste competition a contestant is offered 3 cups
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05 Mar 2013, 08:23
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