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At a blind taste competition a contestant is offered 3 cups of each of
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Updated on: 06 Feb 2019, 23:35
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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. \(\frac{2}{3}\)
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Originally posted by Economist on 14 Nov 2009, 07:34.
Last edited by Bunuel on 06 Feb 2019, 23:35, edited 3 times in total.
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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14 Nov 2009, 08:06
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. 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 (211). \(C^1_3\)  # 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. \(P=1\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1\frac{9}{14}=\frac{5}{14}\). Answer: B. Hope it's clear.
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 Feb 2011, 08:18
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 = \(1  \frac{9}{14} = \frac{5}{14}\)




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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Nov 2009, 04:16
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,
\(P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}\)
\(P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}\)
\(P = \frac{45}{{9*2*7}}\)
\(P = \frac{5}{14}\)



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Aug 2010, 18:43
It should be 1(3*3C2*3C1*3C1/9C4) = 19/14 = 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.
Hope it helps.



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Aug 2010, 19:03
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.
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Aug 2010, 20:06
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?



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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05 Mar 2013, 09:23
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?



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 Mar 2013, 02:32
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: At a blind taste competition a contestant is offered 3 cups of each of
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09 Mar 2013, 22:37
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 of each of
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12 Jan 2015, 08:47
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 of each of
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18 Mar 2016, 00:58
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 of each of
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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 (1opposite), 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



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 May 2017, 10:22
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.



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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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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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Feb 2018, 05:47
To choose 2 types of sample out of 3, we get 3c2 We then have 6 cups and to choose 4 cups out of that we get 6c4. Hence, 3c2 * 6c4 The total ways of selecting 4 cups out of the total 9 is 9c4. Final ans is (3c2 * 6c4) / 9c4
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At a blind taste competition a contestant is offered 3 cups of each of
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22 Feb 2019, 10:49
Can someone explain where I am going wrong?
This was my approach: For the first cup: 9 options For the second cup: 5 options For the third cup: 4 options For the fourth cup: 3 options
(9 * 5 * 4 * 3)/(9C4 * 3!)
I am using the 3! to remove the order



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At a blind taste competition a contestant is offered 3 cups of each of
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22 Feb 2019, 13:20
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. \(\frac{2}{3}\) P(good outcome) = 1  P(bad outcome) Bad outcomes:A bad outcome occurs when all 3 flavors are sampled. For all 3 flavors to be sampled, 1 of the 3 flavors must be chosen TWICE. Number of options for the twicechosen flavor = 3. (Any of the 3 flavors.) From 3 cups of the twicechosen flavor, the number of ways to choose 2 cups = 3C2 = (3*2)/(2*1) = 3. From 3 cups of the next flavor, the number of ways to choose 1 cup = 3. From 3 cups of the last flavor, the number of ways to choose 1 cup = 3. To combine these options, we multiply: 3*3*3*3 All possible outcomes:From 9 cups, the number of ways to choose 4 = 9C4 = \(\frac{9*8*7*6}{4*3*2*1} = 9*2*7\) Thus: P(bad outcome) = \(\frac{badoutcomes}{allpossibleoutcomes} = \frac{3*3*3*3}{9*2*7} = \frac{9}{14}\) P(good outcome) = \(1  \frac{9}{14} = \frac{5}{14}\)
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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05 Jun 2019, 22:21
Although this approach may not be great, I used the elimination method after finding total possible ways of selecting 4 cups out of 9 = 4C9=112.
As I was not able to come up with the exact way to calculate the ways to select 4 cups of 2 samples...i checked for denominators being divisible by 112, as no answer choice can be correct if the denominator isn't a factor of 112. I was left with 1/2 and 5/14 as a potential answer choice, and by gut feeling the case looked to be of probability less than 50%(selecting 4 out of 9) hence excluded 1/2 and marked 5/14.
Though not best, such methods may come handy as a guessing strategy i feel.



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Re: At a blind taste competition a contestant is offered 3 cups of each of
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17 Jun 2019, 06:13
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?




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